What is the derivative of $f(x)= \log^{k}x$ where $k$ is a positive integer?
Update:
Hello I have found that $\log^k n$ = $(\log n)^k$
(Thomas H. Cormen, Introduction to Algorithms (2ed) p. 53 )
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$\begingroup$Use the chain rule (extra characters).
$\endgroup$ $\begingroup$Assuming $k$ is an exponent, use the chain rule, $f(x) = g(h(x))$, $\frac{d f}{d x} = \frac{d g}{d h} \frac{d h}{d x}$. In this case we have $g(h) = h^k$ and $h(x) = \log x$.
$$\frac{d}{dx} (\log x)^k = \frac{d}{d h} \left(h^k\right) \frac{d}{d x} \left(\log x \right) = k h^{k-1} \frac{d}{dx} \log x = k (\log x)^{k-1} \frac{d}{dx} \log x$$
Now, if $\log$ is the natural logarithm you get:
$$\frac{d}{dx} (\log x)^k = \frac{k}{x} (\log x)^{k-1}$$
If $\log$ is the base $b$ logarithm you get $\log x = \frac{\ln x}{\ln b}$ (denoting the natural logarithm by $\ln$):
$$\frac{d}{dx} \log^k x = k (\log x)^{k-1} \frac{d}{dx} \frac{\ln x}{\ln b} = k \frac{(\ln x)^{k-1}}{(\ln b)^{k-1}} \frac{1}{x \ln b} = k \frac{(\ln x)^{k-1}}{x (\ln b)^k}$$
EDIT: Given your source you probably want to use $b = 2$ in the formula above.
EDIT2: Changed notation from $\log^k x$ to $(\log x)^k$ for clarity.
Another way to derive the equation above is to use the product rule $f(x) = f_1(x)f_2(x)$, $f'(x) = f_1'(x)f_2(x) + f_1(x)f_2'(x)$. Since $f(x) = (\log x)^k = \log x \log x \cdots \log x$. Now, when you derive you get one term for every factor in the product, each term containing the derivative once. All the terms are equal so you can simply multiply by the number of terms ($k$) instead:
$$\begin{aligned} \frac{d}{dx} \left(\log x \right)^k = \left(\frac{d}{dx} \log x \right) \log x \cdots \log x \log x + \dots + \log x \log x \cdots \log x \left( \frac{d}{dx} \ \log x \right) = \\ = k \left( \frac{d}{dx} \log x \right) (\log x)^{k-1} \end{aligned}$$
which is the same expression as obtained above.
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