6
$\begingroup$

Hey StackExchange I'm having trouble understating where to start with this problem, I'm supposed to prove something about double derivatives and the chain rule but I'm having trouble understanding exactly what I'm being asked and how to go about doing it.

If $y = f(u)$ and $u = g(x)$, where f and g are twice differentiable functions show that:

$$\frac{d^2y}{dx^2} = \frac{d^2y}{du^2}\left(\frac{du}{dx}\right)^2 + \frac{dy}{du}\left(\frac{d^2u}{dx^2}\right)$$

$\endgroup$
  • $\begingroup$ Start by taking the first derivative of $y$. $\endgroup$ – Code-Guru Jul 12 '14 at 21:23
7
$\begingroup$

Consider $\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{\mathrm{d}f(u)}{\mathrm{d}u}\dfrac{\mathrm{d}u}{\mathrm{d}x}$. Take the second derivative (use the chain rule, product rule, and chain rule, respectively, in that order): $$\dfrac{\mathrm{d}^2y}{\mathrm{d}^2x}=\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{\mathrm{d}f(u)}{\mathrm{d}u}\dfrac{\mathrm{d}u}{\mathrm{d}x}\right)=\dfrac{\mathrm{d}f(u)}{\mathrm{d}u}\dfrac{\mathrm{d}^2u}{\mathrm{d}x^2}+\dfrac{\mathrm{d}^2f(u)}{\mathrm{d}u^2}\left(\dfrac{\mathrm{d}u}{\mathrm{d}x}\right)^2$$

$\endgroup$
3
$\begingroup$

The first derivative $\frac{dy}{dx}$ can be calculated with the chain rule:

$$\frac{dy}{dx}= f'(u)\cdot u' = \frac {dy}{du} \cdot \frac{du}{dx}$$

Now you need to apply the product rule and chain rule to find the second derivative.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.