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

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?

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

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

• You're differentiating $2^k$, not $k^2$. Power rule is not applicable.
– user296602
Commented Oct 4, 2017 at 16:21
• I think @user296602 has hit the nail on the head. +1. Note that there is a useful generalization that encompasses both $x^a$ and $a^x$; if $f$ and $g$ are functions of $x$, and $'$ denotes differentiation with respect to $x$, then $$(f^g)' = gf^{g-1}\cdot f' +f^g\log f\cdot g'$$ If $g$ is constant, you recover the power rule, and if $f$ is constant you recover the exponential rule.
– MPW
Commented Oct 4, 2017 at 16:32
• I don’t quite understand the downvotes. The OP has done everything we request—bring research and explain personal attempts—and I’m sure that plenty of students can learn from this post in the future. Commented Oct 4, 2017 at 17:15

note that $$(a^x)'=a^x\ln(a)$$ and your first derivative is given by $$\left(\frac{2}{3}\right)^{k-1}\left(\ln\left(\frac{2}{3}\right)k+1\right)$$
• You've answered the title, not the body (there was a discrepancy in the title, and it has been fixed.) Should be $(2/3)^{k-1}\cdot k$. Commented Oct 4, 2017 at 16:27
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 $(a^k)'_{k}=a^k\ln{a}$.