Derivative of the summation of the log

Question

Given the function $f(t) = \sum^n_{i=1}{log[(x_i)^{t-1}]}$ where x > 0.

What is the derivative of $f(t)$ with respect to t?

Answer 1

$$f(t) = \sum^n_{i=1}{log[(x_i)^{t-1}]}$$

$$=(t-1)\sum^n_{i=1}{log(x_i)}$$

$$ f'(t) = \sum^n_{i=1}{log(x_i)}$$

Answer 2

Hint :

Use the fact that :

$$\ln[x^t]=t\ln x$$