Derivative of $(2/3)^{k-1}\cdot k$

Question

I want to find the derivative of $f(k)$ but wolfram tells me that it has a $\log$ in it, I have no idea where this $\log$ comes from, did I use the product rule here the wrong?

$f(k)=\left(\frac{2}{3}\right)^{(k-1)}\cdot k$

$f'(k)=(k-1)\cdot\left(\frac{2}{3}\right)^{k-2}\cdot k + \left(\frac{2}{3}\right)^{k-1}$

Answer 1

note that $$(a^x)'=a^x\ln(a)$$ and your first derivative is given by $$\left(\frac{2}{3}\right)^{k-1}\left(\ln\left(\frac{2}{3}\right)k+1\right)$$

Answer 2

It should be: $$f'(k)=\left(\frac{2}{3}\right)^{k-1}k\ln\frac{2}{3} + \left(\frac{2}{3}\right)^{k-1}$$ because $(a^k)'_{k}=a^k\ln{a}$.