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How To Use Implicit differentiation with the quotient rule and trigonometric functions
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Feb 19, 2025
Implicit differentiation is a technique that can be used to find derivatives of functions that are not explicit. The quotient rule states that the derivative of a quotient is equal to the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator. This rule can be used to find derivatives of functions that are expressed in terms of x and y. Trigonometric functions can also be differentiated using implicit differentiation. For example, the derivative of sin x is cos x, and the derivative of cos x is -sin x. These rules can be used together to find derivatives of more complicated functions. For instance, the derivative of sin(x)/x is cos(x)/x-sin(x)/x^2. By using implicit differentiation, it is possible to find derivatives of functions that cannot be easily solved using other methods. In this video, we'll learn how to do that
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0:00
first we need to remember
0:03
the set of rules right so we
0:06
differentiate number one we find the
0:08
derivatives
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in terms of x
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because usually X becomes our
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um
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we do in terms of DX okay if we talked
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about that and the next thing is
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if we collect
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on
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the d y of I over DX terms
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and then three
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we Factor them out
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and then four we just solve
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for d y
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over DX so that's basically the main
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rule
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we find the derivatives in terms of X
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because most problems will call for that
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because sometimes you can do it in terms
1:03
of Y but for the most part we'll be
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dealing with d y over DX
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doesn't mean that we can do DX over d y
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well for the sake of keeping our sanity
1:12
we're gonna do it in terms of X next
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you're going to collect all d y or DX
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terms and then second you put a massive
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Factor and then third and talk you solve
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for d y by DX that's when we do in the
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implicit now we also need to record a
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quotient rule right portion revive a
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function
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h of x
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equals F over G
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what's the derivative H Prime of x value
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[Music]
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um
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forgot yeah I'm sorry so it's F Prime
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Times what G right minus G Prime Times F
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over G squared that's the quotient rule
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okay and then the three functions that
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you may not remember if you don't
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remember that's fine we're probably
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going to have like a cheat sheet and I
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record all those rules so now let's go
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if he has a book
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go to page uh page 146
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and we are going to work on number 12.
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page 146 number 12. so let me write down
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the function
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you guys still need this so you're good
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okay
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146
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number 12.
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so we are basically finding the implicit
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uh differentiation of sine of
2:52
sine pi x
2:54
plus cosine
2:57
by y
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it won't be squared and that's equal
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to 2.
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right
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so sine pi x plus cosine
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Pi y squared equals to 2. so we have to
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use how many rules here 2 right
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we have to use the quotient rules
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and let me know another question rule we
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have to use the chain Rule and then we
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have to also be careful of the
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um what do you call those the trig
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functions so the way we're going to
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solve this is
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I can do it straightforward but I'm not
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going to do that now we're going to try
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and find each one separately and then
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apply the chain rule right remember the
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chain rule when we have U
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to the power of n to find the derivative
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is n u to n minus one
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guys you crying U prime mean the
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derivative of the inside so in this case
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if you were to apply the chain rule gear
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right we will start by finding what two
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times sine
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pi x
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plus cosine Pi y
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2 minus 1 times the derivative of this
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right pi x
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plus cosine of pi y
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Prime okay now we need to find the
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derivative of this we've already done
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the outside so that we need to find the
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derivative of this bracket okay so we're
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doing the implicit remember
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so
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now let's find the derivative
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of
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sine of pi x
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so what's the derivative of sine of 5x
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Eddie
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I'm just messing with you so what's the
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derivative of sine of 5x
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[Music]
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sine of Pi observative
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guys it's the other way around right yes
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first we have to do the inside right pi
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times what
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cosine of pi x that makes sense
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because we finally do you got inside
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which is pi and times cosine of pi x
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okay and now what's the derivative of
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cosine of pi y now remember we're doing
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the implicit so that means it's in terms
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of Y so what would it be
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um a what do you think
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cosine of pi Y is uh
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negative sine of Pi y
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times a cosine of pi
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times just five okay
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just five and then times what
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d y over DX because again we are doing
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the derivative in terms of X okay so is
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pi times cosine of pi x for this
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function and for this function will be
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negative sine Pi y times pi times d y
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over DX yes what what why does the pie
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stay the same throughout the entire
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thing what is the spot here you can put
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it in front of it doesn't matter
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is that what you're asking
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but isn't it
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like like when you find the derivative
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of it like when it equals zero well no
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pi x is the whole thing this is like
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what if I had like sine of 2x right the
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derivative will be 2 cosine of two x pi
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Pi is almost it's a constant it serves
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as a constant just like a two or any
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number would
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so I'm not just finding it separately
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it's the whole expression 2x the
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derivative of 2x is 2 and the derivative
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of pi x is pi all right I'm doing the
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inside and then outside that makes sense
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okay so now we have this
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so now we're going to take all this and
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you know now replace it here okay so now
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we have two
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sine of pi x
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plus cosine of pi y
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times all of these guys add them all
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together good
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so I'm going I'm going to raise this
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because I only I just need to remember
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this so we have
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times
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by
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cosine of pi x
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and then minus
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all these Pi
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sine of
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uh Pi y times d y
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over DX equals to zero the derivative of
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three zero that's gone okay now the the
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next portion is the most not complex
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it's just a lot of work okay so now we
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have to for you so this time that so
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it's going to be
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two
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I'll put Pi cosine of pi x
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times sine of pi x
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plus cosine of my y right minus
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this guy now
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by sine y Pi y times d y over DX
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times this whole thing again sine of pi
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x
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that's cosine with pi y equals zero and
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now we have to solve for d y by DX
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because we just collected the term so
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we're going to move this to the outside
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so that becomes
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negative five sine of Pi y times d y
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over DX times sine of five X plus cosine
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of pi y equals all of that on the other
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side negative 2 pi
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cosine of pi x times sine of pi x plus
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cosine of pi y now I should have seen
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this coming because eventually these are
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going to cancel out so I don't even know
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what I can do anything did you move the
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left hand side equation just off to the
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other side yes I'll move this here
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because we are solving for d y over DX
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that's the end goal right we're trying
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to find the implicit differentiation
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so I'm going to divide this now by
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negative pi
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sine Pi Y and then sine pi x
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plus cosine of pi y we're going to do
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the same thing on both sides so here
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we're going to have negative pi
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sine of
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by y times sine of pi x plus cosine of
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pi Y and eventually these two are going
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to cancel right
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okay so I'm gonna go and then the pies
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are gonna cancel the negatives Are Gonna
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Cancel you're gonna be left with two
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cosine of five x over sine
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y equals to d y
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over DX and that's
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that's the final result does that make
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sense
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it's just very long that's all
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the key here is to make sure you find
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this derivative separately
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and then you can incorporate them back
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into the the problem okay
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foreign
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I should have distributed the two into
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both
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not just one these the two has to
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multiply this guy and the 12 saw so much
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by this guy so we try on the two here as
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well
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we need two
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let me just put two maybe that's two by
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e so that you know if you're looking
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back and then you cancel this
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any questions
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you good
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questions
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but other than that yes well we can do
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that but then you can go back and watch
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that stuff again so that's that's the
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whole thing so we solved one problem
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like between two birds with one stone
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right that's fine the next one we're
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gonna use the quotient rule
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and then I'll let you guys um do some of
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the problem on your own
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now we have X
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x equals one over C over secant
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of Y and then we want to find the
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implicit again differentiation here so
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now we have not only do we have the
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quotient rule but we also have
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um
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what do you call it trig functions but
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do we necessarily have to use the
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quotient rule here do we no right we
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don't necessarily we can turn this into
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a what
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but what can I do to seek one of the
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secret life I can turn this into a
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x equals
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secant y to the negative one
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right and then now we can use the again
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the channel so if you call this X
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equals to secant pi to the negative one
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so now let's find the derivative okay
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we're gonna do the same thing yep okay
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yep
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so we're going to find the derivative so
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what's the derivative of secant
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of Y by itself what would it be
13:00
she could wire the derivative
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[Music]
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secret is secant x tangent X
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right so that would be
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um
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you secant why
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tangent why right times what
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d y with DX right
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that's fine I was really good at
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you got it
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so now now we need to we still need to
13:35
find the derivative of x so what's the
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derivative of x
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what is the derivative
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yeah one all right so one times and then
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we have to apply again the chain rule
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here don't forget that okay so we're
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gonna use the chain rule so it's gonna
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be negative one
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times secant y minus one minus one chain
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rule times the derivative of the inside
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which is all of this
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secant x
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10 chicken y tangent y times d y over DX
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right
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and now this becomes 1 equals negative
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secant y to the negative two times
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secant y
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tangent y d y with the X and then to
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solve for the derivative yeah
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we're using Secret Life remember we're
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doing the derivative implicitly
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so it's in terms of X so you're gonna
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have to keep the d y right here okay if
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you was your second X this should just
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be secant and tangent X without the t y
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like yes
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that makes sense
14:51
all right
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so now we have all of these
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so I'm gonna single out this and I'm
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just gonna divide it right so I'm gonna
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first do to find d y for DX is going to
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be 1 over
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negative
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secant y because every time we divide by
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the same time What Happens we're going
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to divide it by the same time so this is
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gonna we can shift this to the top right
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at some point not right now
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okay
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so we put this behind and then we can
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actually we can work with these two
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together we can combine these two
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together right so you can watch the
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negative two and then second this is
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like to the negative to the positive one
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so we can combine those so you're gonna
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be there for one over
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secant is negative one times tangent y
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equals d y over DX and that just gives
15:59
you d y over DX
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we're going to move this to the top
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change it to into a positive exponent
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algebra 2. so I become secant
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y over tangent y
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okay and that should be your final
16:17
result would it be negative
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negative yeah now what you got
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all right
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questions
16:32
I was I wanted to find one of the
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quotient rules I guess we're gonna do
16:36
23.
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on page 146 23.
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that's like the the quotient rule there
16:50
in China
16:53
so let's do 23 on page 146.
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please
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no I'm not gonna raise it yet
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Back Square minus 49 over x squared plus
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49.
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so basically we're trying to find the
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derivative at a given point we've been
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given the point seven zero
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and we have the function so we're going
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to do the same thing so let's as soon as
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she finishes we'll get to we get to that
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good good all right
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so we've been told that y squared
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is equal to x squared minus 49
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over x squared plus 49
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and we want to they want us to find the
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derivative
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the the slope of a tangent line
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at seven zero slope
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of
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the Vengeance line
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at this specific point so we need to
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find d y over DX right we need to find
18:15
the derivative so we are finding d y
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with DX
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and then once whatever we get there
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we're going to replace X by 7
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and Y by zero to find the
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the slope of the tangent line okay now
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this is a problem where we have to use
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the quotient rule
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so we've just talked about a question
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before we actually started it so we need
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to find in the river so what's the
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derivative of y squared
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two two what
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X so the Y Square by itself sorry 2y
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times d y over DX 2y times d y with DX
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right
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yeah
19:01
you stole the answer right off I'm under
19:04
everyone yeah yeah sometimes anyway so
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let's let's watch the derivative of x
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squared minus 49.
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through
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thus anybody 2x right so 2x times
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this guy right x square plus 49 we use
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the quotient rule
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the minus the derivative of the body
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which is again 2X
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times
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x squared minus 49 this time over x
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squared plus 49 dividing squared right
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now we have to see if you can cancel out
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some like times there and all that so
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we're gonna have two x cubed on top
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plus 49 times 2 that's 98x
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and then negative 2 x cubed and then
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negative 2x times
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negative 49 that's again positive
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98x
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the whole thing over x squared plus 49
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squared and now we can cancel some like
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terms here for now right and then
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commodity 98 into 98
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so we have 2y
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and then times d y over DX
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equals 98 x
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over x squared plus 49 you're holding
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squared now we are not done we're still
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solving for d y over DX so I have to
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divide this whole thing by 2y okay I'm
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going to divide this by 2y
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because we are solving for
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d y over DX
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and what happened to the
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other 98x
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Plus 98x
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thanks for cutting it you combine these
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two together okay
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so
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now we have d y over DX
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equals
21:04
98 plus 98 is equal to what
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that's 196 right
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yeah
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196 x
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over x squared plus 49
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squared revolting over 2y all right
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and then
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again this is like two y over one so
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we're gonna flip this guy so that
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becomes d y
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over the X
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equals 196 x over 2y
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times x squared plus 49
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squared and then we can
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simplify this
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so we have d y
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over DX equals to 98 x
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over y times x squared plus 49. now we
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are done
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all we've done here is find the slope
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right now we need to find the actual
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slope when X is 7 and then Y is zero
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was he zero seven or seven and zero
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let me make sure we do this right
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seven zero yeah
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all right
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well see what's going on here
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huh
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what the what
22:33
the two castles so look we're gonna have
22:36
something over zero that's not good we
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can't have that
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so that means the slow will be only five
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because if I press X by 7 and Y by zero
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we have zero times something that's
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going to be zero so therefore that's the
22:51
problem here but we can't have the slope
22:52
so we'll be undefined does that make
22:54
sense
22:57
yeah
22:59
thank you
23:01
we can investigate this here but there's
23:03
nothing we can do really
23:07
you can't investigate this because this
23:09
is zero times that
23:13
will be undefined in this problem
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that's what we have because I'm I'm
23:19
double checking to see if there's
23:20
anything wrong here but I don't see
23:21
anything wrong with what we did because
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if you divide this by 2y
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and we multiply it I mean we have to
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flip it so yeah so the slope will be
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undefined
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question yes was everything me closing
23:34
being able to be
23:37
squared I guess because
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if you put every put the entire Point
23:42
under a square root it would be y equals
23:45
x minus seven over X plus seven
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if you square root both sides
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would that mean anything
23:54
well
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I mean you're gonna have two I have two
23:57
separate problems you're gonna have two
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separate functions
24:00
we just have you're gonna do them
24:01
separately
24:02
so if you if you were to put let's do it
24:04
so if you were to put this
24:06
right
24:07
you're saying
24:10
if you take the square root of Y squared
24:12
right and then the square root of
24:15
this
24:17
so what you're saying if I'm
24:18
understanding correctly
24:21
so that's going to give you y equals
24:23
plus or minus
24:26
this
24:30
and then you can find the derivative
24:33
x minus seven over X plus seven no this
24:36
is x minus seven times X plus seven this
24:38
is not factual this one here is not
24:40
factual
24:44
it's x squared plus 49. yeah be careful
24:50
the only thing is when you have a square
24:52
minus B squared so this is
24:54
any questions wrong so this is either
24:57
way so the slope is undefined which are
25:00
finding interesting because I really
25:01
thought those huh well it is what it is
25:05
because 2y d y over DX
25:07
either way I should have seen it from
25:09
here look even if we didn't find a
25:11
derivative here you would have seen that
25:13
the derivative will be 2y times d y over
25:15
DX and we would have divided by y anyway
25:17
so that would be undefined does that
25:19
make sense
25:20
all that stuff we did was great but we
25:22
didn't have to because you could
25:24
recognize it from right here right away
25:28
no to her that's fine this is just
25:30
for for the sake of learning how to use
25:32
the quotient rule so that that concludes
25:34
the quotient Rule and implicit
25:36
differentiation and trig functions
25:39
so that should be good and now we can
25:41
work on some
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