# How do you find the derivative of a natural log exponential function y = (ln(x^2))^(2x+3)?

##### 1 Answer
Aug 23, 2015

When we see the variable in both the base and the exponent, we need some form of logarithmic differentiation to find the derivative.

#### Explanation:

$y = {\left(\ln {x}^{2}\right)}^{2 x + 3}$

So

$\ln y = \ln \left({\left(\ln {x}^{2}\right)}^{2 x + 3}\right)$

$\ln y = \left(2 x + 3\right) \ln \left(\ln {x}^{2}\right)$

Differentiate implicitly.

On the left we get: $\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}}$

On the right use the product rule, the derivative of $\ln$ and the chain rule to get some expression $G \left(x\right)$

We'll have: $\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = G \left(x\right)$

so $\frac{\mathrm{dy}}{\mathrm{dx}} = y G \left(x\right)$

$= {\left(\ln {x}^{2}\right)}^{2 x + 3} G \left(x\right)$