0:01
so number one we need to understand how
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to Department an audition right so for
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when you're doing implicitly it's d y
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so for second derivative we use the term
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D squared x so basically it's just a
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change in the way it looks okay the
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two-year indicates second derivative so
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for a third derivative it would just be
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d3y b3x so for the second derivative is
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d y d squared y or d square X
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so if you go to page uh 146 no 147
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so they want you to find the derivative
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plus y squared equals four again this is
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an equation of the circle the circle of
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Center 0 0 radius is two
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so now we want to find the second
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derivative using the implicit so we can
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walk you through the process here
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let's let's go first find the first
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because of the thing you can so you can
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would you solve for d y d x and then get
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the next derivative for a little Chief
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d y over DX equals negative
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X over y okay and then
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we do it again right yeah now what rule
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are we going to use here of course
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question quotient rule right so what
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would it be now we find d square y
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over d square right and
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that's okay or you don't have to
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necessarily use that you can just I mean
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you're gonna have to use the question
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over anyway so let's record it right if
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I have F over G and I want to find the
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derivative I'm doing F Prime G right
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minus G Prime F over G squared okay so
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let's apply that here so we now want to
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so what's the derivative just get the
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negative sign for here now okay
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and let's deal with this what's the
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one one times one one times y minus 1
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times x over and then don't forget that
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oh yeah times d y over DX over
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y squared y squared excellent right now
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we already know what d y over DX is
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we're just going to substitute that here
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so basically you have that so now let's
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go step by step so we have
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I have the value of UI over the years oh
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I'm not going to leave this as p y over
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DX right so negative y minus X right
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y squared right so let's go back here
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so we have negative Y and then
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and negative and a negative you get what
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positive right over y the whole thing
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over y squared now we can put this on
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the same denominator algebra 2.
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so that becomes this and this so that's
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negative y squared right
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plus x squared over y the whole thing
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flip it again Flip or multiply so this
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becomes negative y squared
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plus x squared over y cubed and that's
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going to be your second derivative
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right here I find the same uh
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denominator right I'm gonna put this
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fraction in the same denominator so I'm
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going to want to practice my describe my
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y 'all your lowest common denominator
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okay let's do it here
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wouldn't that just be y over one times
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one over y though uh if I'm putting
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these two together right I wanna this is
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why one right yeah so what's my command
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I'll do the same thing up here
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and then there's another one that we can
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um you guys welcome 48
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48 it's the same thing I'm 48 you have
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and then x minus y thank you
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so in 48 you're gonna have the the the
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when you were walking through this
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so what's the derivative
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here first find d y by DX and then d
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square y over DX so let's start
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so it wasn't derivative of one
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okay so we just know you don't know why
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and now we have a product rule right so
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and then equals on this side
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what's the group of Y
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and now I'm going to open the
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parenthesis so we're going to be
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minus d y over DX times x equals one
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or DX and remember we're going to put
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all the d y over DX's on one side okay
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we're going to collect them
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so I'm going to shift this to this side
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and I'm going to move the Y to the other
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side so that's going to become negative
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and then 1 plus y okay and I'm going to
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factor out the d y over DX so if I put d
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y with DX as a factor
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1 minus x equals 1 plus y
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and now we solve for d y over DX
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that equals one plus y
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1 minus X now this is the first step now
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we have to find d square y over D
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you find the second derivative
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a quotient rule of aiding so what do we
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one plus y why did we wrote These first
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so what's that what's the derivative DX
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it's negative one negative one
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one minus X all squared all right and
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now again let's go back to this we know
8:53
about d y over DX is it's one plus y
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over y minus X so we're going to
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substitute that in there
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so we're going to replace d y over DX
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plus 1 plus X and one minus y
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and something's going to cancel out it's
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going to make our job
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a lot easier now I have two negatives
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that turns equal positive
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these two are going to cancel out
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you're left with one plus y
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over y minus x squared and you just
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combine like terms so basically you have
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with B squared x equals two
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minus x squared so that would be your
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any questions on that
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the x square in the bottom here
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no I don't know why I did that
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that's good that's good