# How do you find the first derivative of y=(lnx)^tanx?

Aug 12, 2016

Take the natural logarithm of both sides.

$\ln \left[y\right] = \ln \left[{\left(\ln x\right)}^{\tan x}\right]$

Simplify the right-hand side using the rule $\ln \left[{a}^{n}\right\} = n \ln a$:

$\ln \left[y\right] = \tan x \left[\ln \left(\ln x\right)\right]$

Differentiate both sides.

$\frac{1}{y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \square$

Inset: $\square$

The derivative of the right hand side is fairly complex. We have to first find the derivative of $\ln \left(\ln x\right)$ using the chain rule, followed by the derivative of $\tan x \times \ln \left(\ln x\right)$, using the product rule.

Let $y = \ln \left(u\right)$ and $u = \ln x$. Then $y ' = \frac{1}{u}$ and $u ' = \frac{1}{x}$.

$u ' = \frac{1}{u} \times \frac{1}{x} = \frac{1}{\ln x} \times \frac{1}{x} = \frac{1}{x \ln x}$

Now we have to find the derivative for $\tan x$ so that we have enough information to apply the product rule.

$\tan x = \sin \frac{x}{\cos} x$

By the quotient rule:

$\left(\tan x\right) ' = \frac{\cos x \times \cos x - \left(- \sin x \times \sin x\right)}{\cos x} ^ 2$

$\left(\tan x\right) ' = \frac{{\cos}^{2} x + {\sin}^{2} x}{{\cos}^{2} x}$

Applying the pythagorean identity ${\cos}^{2} x + {\sin}^{2} x = 1$:

$\left(\tan x\right) ' = \frac{1}{{\cos}^{2} x} = {\sec}^{2} x$

Next, apply the product rule:

Let $y = g \left(x\right) \times h \left(x\right)$, where $g \left(x\right) = \ln \left(\ln x\right) \mathmr{and} h \left(x\right) = \tan x$:

$y ' = \left(\frac{1}{x \ln x} \times \tan x + \ln \left(\ln x\right) \times {\sec}^{2} x\right)$

$y ' = \tan \frac{x}{x \ln x} + {\sec}^{2} x \ln \left(\ln x\right)$

Putting this into the place of the $\square$:

$\frac{1}{y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \tan \frac{x}{x \ln x} + {\sec}^{2} x \ln \left(\ln x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\tan \frac{x}{x \ln x} + {\sec}^{2} x \ln \left(\ln x\right)}{\frac{1}{y}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = y \left(\tan \frac{x}{x \ln x} + {\sec}^{2} x \ln \left(\ln x\right)\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(\ln x\right)}^{\tan x} \left(\tan \frac{x}{x \ln x} + {\sec}^{2} x \ln \left(\ln x\right)\right)$

I know it's quite messy, but it worked!!

Practice exercises:

1. Determine the derivative of the given relation.

a) $f \left(x\right) = {\log}_{2} \left(\sin \left(3 x + 4\right)\right)$

b) $g \left(x\right) = {\left(2 \ln x\right)}^{\sec \left(3 x + 5\right)}$

Hopefully this helps, and good luck!