# Calculus AB/BC – Implicit Differentiation hi everyone today we’re going to be
covering implicit differentiation and this is the first in my series of
calculus videos because I had a lot of requests I decided to calculus instead of algebra II but I’ll do algebra II later on and if you have any questions you can still ask. So implicit
differentiation. So you use implicit differentiation when differentiation when it’s hard to solve
the equation for y but we have to find dy/dx. So I’m gonna give an
example here. 4x^2 -9 y^2=17 This is my example.
If we solve this equation for y we can still do it — we can add 9y^2
to both sides, but then we’d have to square or square root the whole thing and then
to differentiate a square root, it’s going to take a long time so it’s a lot
easier to use this thing called implicit differentiation. So what we do when
you want do implicit differentiation is that we first take
the derivative of everything with respect to x. So I’m just going to write that down — so the
derivative of 4x^2 squared with respect to x minus the derivative with respect to x of
9y^2 and the derivative with respect to x of 17. Okay and then we’re just going to solve this. So the derivative of 4x^2 is just going to be 8x, because you move the two over and you do 2*4, and then x^(2-1) equals x^1 but then here it’s going to be something different and
what we’re gonna do is we’re going to take the derivative with respect to Y of
9y^2 first and then we’re going to do the chain rule. When we do the chain rule, we take the derivative of what’s inside, so basically dy/dx. So if you don’t understand I’m going to try to explain it a little bit more so what we’re assuming when we see
y is that y is a function of x, so y is a function of x. And then when we take the derivative of this with respect to x if you remember from the chain rule, we take the derivative with respect to y of y(x) times the derivative of y(x), which
is dy/dx, if you remember that from the chain rule. Ok I’ll explain it another way as well but
what we see here is not y(x), we see 9y(x)^2 so we’re taking the derivative
with respect to x of 9y(x)^2 we’re just taking this part and we’re
squaring that. And when we do that with the chain rule we take the derivative with
respect to x, oh sorry, with respect to Y of the whole thing, times what’s inside. That’s what I wrote here, except I just didn’t write the y(x). So I hope you understand this part and when we take the derivative of 17, it’s just going to be 0. And then I’m going to simplify this a
bit further so we have 8x minus And when we take the derivative with respect to y of 9y^2, we have 18y. times dy/dx, equals 0. And then we can add 18y*dy/dx to both sides. So we have 8x=18y times dy/dx. And then we can divide 18
y to get dy/dx equals 8x over 18y and if we divide by 2 we have
4x/9. So that’s our answer. 4x/9y.
So that’s how we do implicit differentiation. We’re basically taking the derivative with respect to x of the
whole equation and then solving for dy/dx. I’m gonna try a few more examples, oh and by the way, hold on so if you practice these kind of problems a lot then you
can skip this writing writing like all these derivative with
respect to x so you don’t really need to write that every step if you can
skip steps I guess I mean it’s always nice to write out you work, but it’s faster if
you just go to this step right away you know? But for the first few videos — I mean the first few questions — I’m going to write everything
out, but then after a while I think I’m gonna stop writing it out. But I’ll tell you when I stop writing it out. So first we’re going to take the derivative with respect to x of
each thing. So x^3 + the derivative with respect to x of y^3=the derivative with respect to x of xy. So this is just going to be 3x^2 + and here we have the Y again so we need
to take the derivative with respect to dy of Y^3 times dy/dx.
equals and here when we take the derivative of this we are going to use the product
rule. So we have the first, which is x, times the derivative of the second which is, and since this is going to be a y, we have to do the derivative of Y with respect to Y times dy/dx. Plus the second times the derivative of the first. The derivative of X is just one. I’m just going to simplify this a bit, so 3x^2 + the derivative of y^3 with respect to Y is just
3y^2 times dy/dx, equals… and here we just have the X, so we’re going to take the derivative of Y with respect to Y, which is just 1, so it’s 1 times dy/dx, which is just dy/dx, + Y. And then we want to put the dy/dx on one side so that we can solve for dy/dx. I’m going to move this 3x^2
to the right side by subtracting it, so we have -3x^2 and I’m going to keep this Y here as well.
And then we’re going to keep this on the left side so 3y^2 (dy/dx). And then we’re going to subtract it by this. So -x *(dy/dx), so now have the dy/dx on both sides so we can do dy/dx times 3y^2 – x, and factor the dy/dx out. equals – 3x^2+y, and then all we have to do is divide 3y^2-x on both sides so we get dy/dx=-3x^2+y over 3y^2-x. Let me check the answer. Okay it’s all good. everything here but we can also just skip to this step as I told you before and I’m gonna start doing that but I’ll explain it on the side, like why I’m going to do each step so So then for this question, since we
already practiced a few, you’re going to pause it and try it on your own and then check
your answer in the end ok so I’m assuming you paused it, and if you didn’t that’s okay as well. So I’m gonna take the derivative with respect to x for the
whole equation The derivative with respect to x of y=the derivative with respect to x of e^(xy). So when we do this, remember that we have to take the
derivative of Y with respect to Y and then multiply with dy/dx, so the derivative
of Y with respect to Y is just 1, so it’s going to be 1 times dy/dx equals and for this we’re just going to take the derivative of this. So we have e^(xy), and this is chain rule so we have to take the derivative of xy as well. So X times dy/dx, because remember when we take the derivative of Y it’s going to be the
derivative of Y with respect to Y times dy/dx. Plus Y times the derivative of the first value, x, which is just going to be 1. Then we have dy/dx=x*e^(xy)*(dy/dx) + y*e^(xy) And then we want dy/dx on one side of the equation, so I’m going to subtract this x*e^(xy), so we have dy/dx x*e^(xy)*(dy/dx)=ye^(xy) And then we can factor the dy/dx out, so we get (1-x*e^*(xy))=y*e^(xy). Then we have dy/dx=y*e^(xy), and then we’re just going to divide this (1-x*e^(xy)) so we have (1-x*e^(xy)) Let me check if the answer is right. So this is going to involve a tangent, and when we have trigonometry involved, we still use the same trig rules that we had before. So the
derivative of x with respect to x is just 1=going to take the derivative of the
whole thing, but then remember to do the chain rule of the inside. So the derivative of tangent — or the trig function rule for that is secant^2. So we have sec^2(x+y), but then we need to multiply that with the derivative of the inside. So we have the first times the derivative – oh, just kidding. This is not the product rule, this is just adding. Ok, so the derivative of x with respect to x is just 1+ the derivative of y with respect to x, we remember to first do the derivative with y respect to y, which is just 1, times dy/dx. And then we’re just going to simplify this. So 1=sec^2(x+y) + sec^2(x+y)*(dy/dx) So I’m just doing this times this, plus this times this. And then I’m going to move sec^2(x+y) to the left by subtracting it, so we have 1- sec^2(x+y)=sec^2(x+y) *(dy/dx). ok and if you
remember trig identities — I’m gonna write it on the side right here, but we have
tan^2x+1=sec^2x So if you remember that, if we have 1-sec^2(x), this is just going to be -tan^2x. So we can change this to -tan^2(x+y) and simplify it like that. Then we can divide by sec^2(x+y), so we have dy/dx=-tan^2(x+y) all over sec^2(x+y) And we can simplify this further, because we know that tan=sin/cos and sec=1/cos, so for tangent, we have -sin^2(x+y)/cos^2(x+y) this is tan^2(x+y) right? But we’re dividing that by sec^2(x+y) which is the same thing as multiplying
it by the reciprocal of sec^2(x+y) so since sec^2(x+y)=one over cos^2(x+y), we’re just going to multiply this with cos^2(x+y), and then we can cancel out the cos^2(x+y), and we’re left with -sin^2(x+y) Ok, so this is our answer. ok so hope you enjoyed watching the
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