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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?

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

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

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    $\begingroup$ You're differentiating $2^k$, not $k^2$. Power rule is not applicable. $\endgroup$
    – user296602
    Commented Oct 4, 2017 at 16:21
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    $\begingroup$ 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. $\endgroup$
    – MPW
    Commented Oct 4, 2017 at 16:32
  • $\begingroup$ 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. $\endgroup$
    – gen-ℤ ready to perish
    Commented Oct 4, 2017 at 17:15

2 Answers 2

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

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  • $\begingroup$ this just slept out of my mind, feel bad for asking now ... $\endgroup$
    – Oleg
    Commented Oct 4, 2017 at 16:24
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    $\begingroup$ You've answered the title, not the body (there was a discrepancy in the title, and it has been fixed.) Should be $(2/3)^{k-1}\cdot k$. $\endgroup$
    – Thomas Andrews
    Commented Oct 4, 2017 at 16:27
  • $\begingroup$ ok thank you for the hint, it is just corrected, important such a difference $\endgroup$
    – Dr. Sonnhard Graubner
    Commented Oct 4, 2017 at 16:31
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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 $(a^k)'_{k}=a^k\ln{a}$.

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