I understand how to do differentiation when the elements are individual functions, but am having trouble applying the concepts with nested functions.

For example, given $(\cos \pi x + \sin \pi y)^5= 24$ $$\frac{d}{dx}\left[\frac{d}{dx}(\cos \pi x)+\frac{d}{dx} (\sin \pi y)\right]^5=\frac{d}{dx}24$$

Following the chain rule of ${f}'(g(x))\cdot {g}'(x), $

$$5[(\cos(\pi x) +\sin (\pi y)]^4 \cdot [-\pi \sin(\pi x) + \pi \cos(\pi y)]=0$$

That's where I get stuck. First, as I see it, I have a few options for the next step:

  • multiply ${g}'(x)$ by 5

  • multiply $g(x)$ by 4

What about after that? How do I separate the contents inside the power block to move the $x$ functions to the other side of the equation?

  • $\begingroup$ Also, your notation is not correct here, although you did take the derivative properly, with the exception that pointed out below. $\endgroup$ – JavaMan Mar 4 '11 at 16:44
  • $\begingroup$ Do you want to find $dy/dx$, given $(\cos \pi x + \sin \pi y)^5= 24$? $\endgroup$ – Américo Tavares Mar 4 '11 at 19:28
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    $\begingroup$ If $F(x,y)=0$ and $y=f(x)$, then $\dfrac{\partial F}{\partial x}+\dfrac{% \partial F}{\partial y}\dfrac{dy}{dx}=0$, $\dfrac{dy}{dx}=-\dfrac{\dfrac{% \partial F}{\partial x}}{\dfrac{\partial F}{\partial y}}$ $\endgroup$ – Américo Tavares Mar 4 '11 at 20:20

Jason. You should not be solving for $y$, you should be solving for $\frac{dy}{dx}$. There was an error in your calculation, however, when you applied the chain rule to find $\frac{d}{dx} (\sin (\pi y))$. It should be:

$$ \frac{d}{dx} \left( \sin ( \pi y) \right) = \cos(\pi y) \frac{d}{dx} (\pi y) = \cos (\pi y) \pi \frac{dy}{dx}, $$

since we are viewing $y$ as a function of $x$. Try now to solve for $\frac{dy}{dx}$ with what you've been given here.

  • $\begingroup$ This is what made it work, since that last$\frac{dy}{dx}$ is what is left over for that side of the equation once everything is shifted over. I missed that part in the prof's examples. $\endgroup$ – Jason Mar 4 '11 at 22:44

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