# How do you find the derivative of log_5(arctanx^x)?

Jul 16, 2017

$\left(\text{d")/("d} x\right) {\log}_{5} \left(\arctan \left({x}^{x}\right)\right) = \frac{\left(1 + \ln \left(x\right)\right) {x}^{x}}{\ln \left(5\right) \left({x}^{2 x} + 1\right) \arctan \left({x}^{x}\right)}$.

#### Explanation:

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

The right hand side of this expression is fairly horrible. It would be better to rearrange this and use implicit differentiation than attempt to differentiate directly.

${5}^{y} = \arctan \left({x}^{x}\right)$.

A common trick to help differentiate functions involving exponents is to write the base as $e$ and modify the power with a logarithm to give the desired exponent. This then makes differentiating easier. Notice that ${x}^{x} = {e}^{x \ln \left(x\right)}$ and ${5}^{y} = {e}^{y \ln \left(5\right)}$. Then,

${e}^{y \ln \left(5\right)} = \arctan \left({e}^{x \ln \left(x\right)}\right)$.

Differentiate implicitly. Note that "d"/("d"x) arctan(x) = 1/(x^2+1).

$\ln \left(5\right) {e}^{y \ln \left(5\right)} \left(\text{d"y)/("d"x)=1/(e^{2xln(x)}+1) "d"/("d} x\right) \left({e}^{x \ln \left(x\right)}\right)$.

As "d"/("d"x) e^{xln(x)} = (1+ln(x))e^{xln(x)}, we see that,

$\left(\text{d"y)/("d} x\right) \ln \left(5\right) {e}^{y \ln \left(5\right)} = \frac{1}{{e}^{2 x \ln \left(x\right)} + 1} \left(1 + \ln \left(x\right)\right) {e}^{x \ln \left(x\right)}$.

We can now rewrite the exponents as what they were originally and rearrange for $\left(\text{d"y)/("d} x\right)$.

$\left(\text{d"y)/("d} x\right) = \frac{\left(1 + \ln \left(x\right)\right) {x}^{x}}{\left({x}^{2 x} + 1\right) \ln \left(5\right) {5}^{y}}$.

We know from the definition that ${5}^{y} = \arctan \left({x}^{x}\right)$.

We conclude,

$\left(\text{d"y)/("d} x\right) = \frac{\left(1 + \ln \left(x\right)\right) {x}^{x}}{\ln \left(5\right) \left({x}^{2 x} + 1\right) \arctan \left({x}^{x}\right)}$.