$\log(y)=a+b\log(1+f(x)) $. Is $\frac{d\log(y)}{d\log(x)}= \frac{d\log\left(y\right)}{d\log\left(x\right)}=\frac{b}{1+f\left(x\right)}\times f'(x)$?

Question

Imagine that we have the following logarithmic function: $$ \log\left(y\right)=a+b\log\left(1+f\left(x\right)\right) $$ I wish to compute $\frac{d\log\left(y\right)}{d\log\left(x\right)}.$ Would this be: $$ \frac{d\log\left(y\right)}{d\log\left(x\right)}=\frac{b}{1+f\left(x\right)}\times f'(x) $$

Thanks!

Answer 1

Another way: $$\dfrac{d\log(y)}{dx}=\frac{d\log\left(y\right)}{d\log\left(x\right)}\dfrac{d\log(x)}{dx}=\frac{d\log\left(y\right)}{d\log\left(x\right)}\dfrac{1}{x}$$

$$\frac{d\log\left(y\right)}{d\log\left(x\right)}=x\dfrac{d\log(y)}{dx}=x\frac{bf'(x)}{1+f\left(x\right)}$$