## Study23 Group Title Implicit Differentiation, please help! $$\ \large cos(x-y)=xe^x .$$ one year ago one year ago

1. freewilly922

Implicit D, is basically the chain rule without finishing. In general you have something like $f(x) = (\text{stuff})^5$ and $f'(x) =5(\text{stuff})^4 (\text{stuff})^{\prime}$ Well you do the same thing here except when you have to do something like $\frac{d}{dx} y^2= 2y y'$ So for example Implicit D of $\sin\bigg(xy^2\bigg)$ give $\cos\bigg(xy^2\bigg)\cdot\bigg( y^2 + x\cdot(2yy^{\prime})\bigg)$

2. Study23

What I'm not understanding is that cos(x-y) part. How do I take the derivative of that??

3. freewilly922

If you substituted u(x) for (x-y) you would have $\frac{d}{dx} \cos(u(x))$ which according to the chain rule would be $-\sin(u(x)\cdot (u'(x))$ do you follow that?

4. freewilly922

Sorry that should be $-sin(u(x))\cdot (u'(x))$

5. Study23

So the two function we have here are cos x and x-y? So we would use the chain rule?

6. freewilly922

yes. the only thing is that when you take the derivative of u(x) you are taking the derivative of "x-y" with respect to x. So you would have $\frac{d}{dx}x + \frac{d}{dx}y$ which becomes $1 + \frac{d}{dx}y$ or $1 + y'$ The fact that you don't finish taking the derivative of y, because you don't know what it is is the "unfinished" part of the chain rule.

7. freewilly922

crap, put a minus sign there in front of the y and y prime

8. Study23

So for the derivative of x-y, with would be 1- y dy/dx ? Why doesn't the y become a one too?

9. freewilly922

no it would be 1 -dy/dx you're instincts are correct

10. freewilly922

$\frac{d}{dx}(x-y) \to\frac{d}{dx}(x)+\frac{d}{dx}(-y) \to 1+y'$

11. Study23

Okay, so what should the correct answer be? So I can check if I did this correctly??

12. freewilly922

So for an exponential function $\frac{d}{dx} e^{x^2y^3}\to e^{x^2y^3}\frac{d}{dx}(x^2y^3)$ and so on

13. Study23

Okay. Thanks!

14. freewilly922

I've been explicitly warned not to give answers to problems directly. The best I can do is give similar examples and hope that it helps. The general process though is to implicitly differentiate both sides and then collect and isolate the y'. so if you had $x+y' = \sin(x) + xy'$ you would isolate the terms with y prime, factor y prime out and then divide by the non y prime factor. $x+y' -\sin(x)= -y' + xy'$ $x-\sin(x)= y'(-1 + x)$ $\frac{x-sin(x)}{1-x}=y'$ notice for this problem you will not get a y prime on the RHS since there is no y variable in that function.