2
$\begingroup$

I saw this from these online notes about differential equations.

$\frac{\mu'(t)}{\mu(t)} = p(t)$

Hopefully you will recognize the left side of this from your Calculus I class as the following derivative.

($\ln\mu(t))' = p(t)$

I don't think I understand why $\frac{\mu'(t)}{\mu(t)} = (\ln\mu(t))'$ is true. Why is this the case? Thank you very much!

|cite|improve this question
$\endgroup$
  • 3
    $\begingroup$ Two words: chain rule. $\endgroup$ – Sean Roberson Aug 4 '18 at 22:32
  • $\begingroup$ Recall that the chain rule yields that $\frac{d}{dt} f(g(t)) = f'(g(t))*g'(t)$. In your case, we'll call $f(t) = \text{ln} (t)$ and let $g(t) = \mu (t)$ $\endgroup$ – Theo C. Aug 4 '18 at 22:32
  • $\begingroup$ Oh okay, that makes perfect sense now! I forgot to try the chain rule. Thank you. $\endgroup$ – Beneschan Aug 4 '18 at 22:34
3
$\begingroup$

It is the chain rule $$(\ln(f(x))'=\frac{1}{f(x)}\cdot f'(x)$$ for $$f(x)>0$$

|cite|improve this answer
$\endgroup$
0
$\begingroup$

Apply chain rule : $$(\ln( u(t))'=\frac {d}{dt} \ln (u(t))=\frac {d \ln u}{du}\frac {du}{dt}=\frac 1 {u(t)}\frac {du}{dt}=\frac {u'}{u}$$

|cite|improve this answer
$\endgroup$
0
$\begingroup$

If we let $$I=\int\frac{f'(x)}{f(x)}dx$$ then we can use substitution $u=f(x) \therefore dx=\frac{du}{f'(x)}$ so the integral becomes: $$I=\int\frac{1}{u}du=ln|u|=ln|f(x)|+C \therefore \frac{d}{dx}\left[\ln(f(x))\right]=\frac{f'(x)}{f(x)}$$

|cite|improve this answer
$\endgroup$

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.