# How do I find the derivative of y=tanx^secx + secx^cotx?

May 31, 2017

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

#### Explanation:

Let $y = \tan {x}^{\sec} x + \sec {x}^{\cot} x$

Let $u = \tan {x}^{\sec} x$ and $v = \sec {x}^{\cot} x$

$\ln u = \sec x \ln \tan x$

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

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

$\ln v = \cot \ln \sec x$

$\frac{1}{v}$ $\left(v '\right) = - {\csc}^{2} x \ln \sec x + 1$

$v ' = v \left(1 - {\csc}^{2} x \ln \sec x\right) = \left(1 - {\csc}^{2} x \ln \sec x\right) \sec {x}^{\cot} x$

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