Partial Derivative of $(b-1) \ln(1-x^a)$ with respect to $a$

Question

What is the derivative of $(b-1)\ln(1-x^a)$ with respect to $a$?

I found that answer will be $\frac{-(b-1) x^a\ln(x)}{1-x^a}$.

But I do not understand why there is $\ln x$ in the partial derivation.

Please help.

Answer 1

We can easily differentiate until this point: $$\frac\partial{\partial a}(b-1)\ln(1-x^a)=(b-1)\cdot\frac1{1-x^a}\cdot\frac\partial{\partial a}(1-x^a)$$ But what is the last term? We rewrite $x^a=e^{a\ln x}$, which differentiates to $e^{a\ln x}\ln x=x^a\ln x$, and this is where $\ln x$ comes from.

Answer 2

$f(a) =(b-1)\ln(1-x^{a})$, then $f(a) = (b-1)\ln(1-e^{a\ln(x)})$

Well $e^{a\ln(x)} = (e^{\ln(x)})^{a} = x^{a}$, then $( x^{a})' = (e^{a\ln(x)})' = ln(x)e^{a\ln(x)}=\ln(x)x^{a}$

So $f'(a) = (b-1) \frac{-\ln(x)x^{a}}{1-x^{a}}$