Can chain rule be used in first step

Question

I was wondering if it were possible to use the chain rule in the first step to differentiate the f.f.g function:

$$f(x) = (1 + \sin x)^{\cot x}$$

I know the obvious first step is to use the power rule but out of curiosity, could the chain rule be applied in the first step?

Answer 1

$\ln{f(x)}=cot(x)\ln(1+sin\,x)$ $$\frac{f'(x)}{f(x)}=-(1+cot^2x)\ln(1+sin\,x)+\frac{cos\,x}{1+sin\,x}cot\,x$$ then $$f'(x)=\left[-(1+cot^2x)\ln(1+sin\,x)+\frac{cos\,x}{1+sin\,x}cot\,x\right](1+sin\,x)^{cot\,x}$$

Answer 2

Whenever you need to compute the derivative of an expression which only contains products, quotients and powers, logarithmic differentiation makes life very simple.

Consider for example $$y=\frac{f(x)^a g(x)^b}{ h(x)^{c}}$$ where $a,b,c$ are constants. Take logarithms $$\log(y)=a \log(f(x))+b \log(g(x))-c \log(h(x))$$ Now $$\frac{y'}y=a \frac{f'}f+b\frac{g'}g-c\frac{h'}h$$ and then $y'$.

In a case similar to your $$y=f(x)^{g(x)}$$ $$\log(y)=g(x) \log(f(x))$$ and apply the product rule; so $$\frac{y'}y=g' \log(f)+g \frac {f'}f$$ and then $y'$.