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

Question

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!

Answer 1

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

Answer 2

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

Answer 3

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: $$I=\int\frac{1}{u}du=ln|u|=ln|f(x)|+C \therefore \frac{d}{dx}\left[\ln(f(x))\right]=\frac{f'(x)}{f(x)}$$