# Why is u'(t)/u(t) = (ln(u(t)))'?

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!

• Two words: chain rule. – Sean Roberson Aug 4 '18 at 22:32
• 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 $g(t) = \mu (t)$ – Theo C. Aug 4 '18 at 22:32
• Oh okay, that makes perfect sense now! I forgot to try the chain rule. Thank you. – Beneschan Aug 4 '18 at 22:34

It is the chain rule $$(\ln(f(x))'=\frac{1}{f(x)}\cdot f'(x)$$ for $$f(x)>0$$
Apply chain rule : $$(\ln( u(t))'=\frac {d}{dt} \ln (u(t))=\frac {d \ln u}{du}\frac {du}{dt}=\frac 1 {u(t)}\frac {du}{dt}=\frac {u'}{u}$$
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: $$I=\int\frac{1}{u}du=ln|u|=ln|f(x)|+C \therefore \frac{d}{dx}\left[\ln(f(x))\right]=\frac{f'(x)}{f(x)}$$