# How do you differentiate y= x^(11cosx)?

May 30, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = 11 {x}^{11 \cos \left(x\right)} \left(\cos \frac{x}{x} - \sin \left(x\right) \ln \left(x\right)\right)$

#### Explanation:

$y = {x}^{11 \cos \left(x\right)}$

To deal with tricky exponents like this, let's take the natural logarithm of both sides and remember the rule $\log \left({a}^{b}\right) = b \log \left(a\right)$.

$\ln \left(y\right) = \ln \left({x}^{11 \cos \left(x\right)}\right)$

$\ln \left(y\right) = 11 \cos \left(x\right) \ln \left(x\right)$

Now take the derivative on both sides. On the left, we'll need the chain rule. On the right, we'll use the product rule.

$\frac{1}{y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 11 \left(\frac{d}{\mathrm{dx}} \cos \left(x\right)\right) \ln \left(x\right) + 11 \cos \left(x\right) \left(\frac{d}{\mathrm{dx}} \ln \left(x\right)\right)$

$\frac{1}{y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 11 \left(- \sin \left(x\right)\right) \ln \left(x\right) + 11 \cos \left(x\right) \left(\frac{1}{x}\right)$

Solving for the derivative:

$\frac{\mathrm{dy}}{\mathrm{dx}} = y \left(\frac{11 \cos \left(x\right)}{x} - 11 \sin \left(x\right) \ln \left(x\right)\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 11 {x}^{11 \cos \left(x\right)} \left(\cos \frac{x}{x} - \sin \left(x\right) \ln \left(x\right)\right)$