0
$\begingroup$

Suppose I have the following function where $$z=\omega(\zeta)=\frac{1}{\zeta}$$ and also, $$\phi(\zeta) = \zeta^{-1}+2\zeta$$ By using chain rule, I can get the first-derivative of $\phi(z)$. Notice that I want now $\phi$ as a function of $z$. Thus, $$\phi'(z)=\frac{d\phi}{d\zeta}\frac{d\zeta}{dz}=\frac{\phi'(\zeta)}{\omega'(\zeta)}$$ where, $$\omega'(\zeta) = \frac{d\omega(\zeta)}{d\zeta}=\frac{dz}{d\zeta}$$

My question is, how can we get the second-derivative of $\phi(z)$, i.e. $\phi''(z)$? By using chain rule?

$\endgroup$
0
$\begingroup$

You already have $\phi'(z)$, so just differentiate it using the product and chain rules:

$$\phi''(z) = \frac{d}{dz}\left(\frac{d\phi}{d\zeta}\right)\frac{d\zeta}{dz} + \frac{d\phi}{d\zeta}\frac{d}{dz}\left(\frac{d\zeta}{dz}\right) = \frac{d^2\phi}{d\zeta^2}\left(\frac{d\zeta}{dz}\right)^2+\frac{d\phi}{d\zeta}\frac{d^2\zeta}{dz^2}.$$

Notice that nothing we've done uses the form of your functions, so this holds in full generality. In your special case, you could also just substitute everything into get $\phi$ in terms of $z$ directly, then differentiate that twice.

$\endgroup$
  • $\begingroup$ Thanks @user3482749! I am actually trying to avoid direct substitution but rather go with this implicitly through chain rule. $\endgroup$ – BeeTiau Nov 15 '18 at 18:00
  • $\begingroup$ I still don't understand how to get the first part in the last expression, i.e. $\frac{d^2\phi}{d\zeta^2}\left(\frac{d\zeta}{dz}\right)^2$ $\endgroup$ – BeeTiau Nov 15 '18 at 18:08
  • $\begingroup$ Sure: $\frac{d}{dz}\left(\frac{d\phi}{d\zeta}\right) = \frac{d^2\phi}{d\zeta^2}\frac{d\zeta}{dz}$ by the chain rule (applied to the function $\frac{d\phi}{d\zeta}$), then it's multiplied by $\frac{d\zeta}{dz}$, so I've grouped the two copies of $\frac{d\zeta}{dz}$ together. $\endgroup$ – user3482749 Nov 15 '18 at 18:09

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.