# How do you find the derivative of x^(log(base5)(x))?

Sep 18, 2016

$2 {x}^{{\log}_{5} \left(x\right) - 1} {\log}_{5} \left(x\right)$

#### Explanation:

$y = {x}^{{\log}_{5} \left(x\right)}$

Take the natural logarithm of both sides. (This is known as logarithmic differentiation.)

$\ln \left(y\right) = \ln \left({x}^{{\log}_{5} \left(x\right)}\right)$

Use the rule: $\log \left({a}^{b}\right) = b \cdot \log \left(a\right)$ to rewrite this function:

$\ln \left(y\right) = {\log}_{5} \left(x\right) \cdot \ln \left(x\right)$

Now, rewrite ${\log}_{5} \left(x\right)$ in logarithms with base $e$ using the change of base formula: ${\log}_{a} \left(b\right) = {\log}_{c} \frac{b}{\log} _ c \left(a\right)$.

$\ln \left(y\right) = \ln \frac{x}{\ln} \left(5\right) \cdot \ln \left(x\right)$

$\ln \left(y\right) = {\left(\ln \left(x\right)\right)}^{2} / \ln \left(5\right)$

Differentiate both sides. The chain rule will be needed on both sides of the equation.

$\frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \ln \left(x\right)}{\ln} \left(5\right) \cdot \frac{d}{\mathrm{dx}} \ln \left(x\right)$

We already have used the derivative of $\ln \left(x\right)$ is $\frac{1}{x}$. Also, rewrite $y$ as ${x}^{{\log}_{5} \left(x\right)}$:

$\frac{1}{x} ^ \left({\log}_{5} \left(x\right)\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \ln \left(x\right)}{x \ln \left(5\right)}$

Note that $\ln \frac{x}{\ln} \left(5\right) = {\log}_{5} \left(x\right)$ through the change of base formula in reverse, or ${\log}_{c} \frac{b}{\log} _ c \left(a\right) = {\log}_{a} \left(b\right)$.

$\frac{1}{x} ^ \left({\log}_{5} \left(x\right)\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 {\log}_{5} \left(x\right)}{x}$

Multiplying both sides by ${x}^{{\log}_{5} \left(x\right)}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 {x}^{{\log}_{5} \left(x\right)} {\log}_{5} \left(x\right)}{x}$

Note that ${x}^{{\log}_{5} \left(x\right)} / x = {x}^{{\log}_{5} \left(x\right) - 1}$ through the rule that ${x}^{a} / {x}^{b} = {x}^{a - b}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 {x}^{{\log}_{5} \left(x\right) - 1} {\log}_{5} \left(x\right)$