Derivative and chain rule exam problem

Question

I can't understand what should I do here? I'm gonna have this problem on the exam (a similar one). Can you guys help me to understand what to do?

Show this formula ${d\over dx}$${f(x)g(x)=f'(x)g(x)+f(x)g'(x)}$ , by finding derivative of ${\ln (f(x)g(x))}$ with help of chain rule.

You are allowed to use rules like this ${d\over dx}$$({f(x)+ kg(x))=f'(x)+kg'(x)}$

Answer 1

$$(f(x)g(x))'\frac{1}{f(x)g(x)}\color{red}=\frac{d}{dx}(\ln (f(x)g(x)))=\frac{d}{dx}(\ln (f(x))+\ln (g(x)))=$$ $$\frac{d}{dx}\ln (f(x))+\frac{d}{dx}\ln (g(x))\color{red}=f'(x)\frac{1}{f(x)}+g'(x)\frac{1}{g(x)}=\frac{f'(x)g(x)+f(x)g'(x)}{f(x)g(x)}$$

by using the chain rule in the $\color{red}{\text{red equalities}}$.

Now multiply by $f(x)g(x)$ to obtain $$(f(x)g(x))'=f'(x)g(x)+f(x)g'(x)$$

Answer 2

Hint: $\ln(ab) = \ln(a) + \ln(b)$

Answer 3

From the Chain Rule, $$\frac d{dx} (fg)=\frac{\partial f}{\partial fg}\cdot\frac{df}{dx}+\frac{\partial g}{\partial fg}\cdot\frac{dg}{dx}=g\cdot f'+f\cdot g'$$ so $$\frac{d}{dx}f(x)g(x)=f'(x)g(x)+f(x)g'(x)$$ which is the Product Rule.

Hence $$(\ln fg)'=\frac d{dfg}\ln (fg)\cdot\frac{d}{dx}fg=\frac1{fg}\cdot(g\cdot f'+f\cdot g')$$ In other words, $$\boxed{\frac d{dx}\ln(f(x)g(x))=\frac{1}{f(x)g(x)}\left(f'(x)g(x)+f(x)g'(x)\right)}$$