How do you find the derivative of ${cos x}^{tan x}$ $cosx^tanx$ ?

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How do you find the derivative of ${cos x}^{tan x}$ $cosx^tanx$ ?

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Answer 1 · 2017-07-16T15:12:33

$\frac{d}{d x} ({cos x}^{tan x}) = ({sec}^{2} x ln cos x - {tan}^{2} x) {cos x}^{tan x}$ $d/dx(cosx^tanx)=(sec^2xlncosx-tan^2x)cosx^tanx$

Explanation:

First define $y = {cos x}^{tan x}$ $y= cosx^tanx$

Then, by definition, $ln y = tan x ln cos x$ $lny =tanxlncosx$

And $\frac{1}{y} (\frac{d y}{d x}) = {sec}^{2} x ln cos x - (\frac{sin x}{cos x}) tan x$ $1/y(dy/dx) = sec^2 xlncosx -(sinx/cosx)tanx$

$\frac{d y}{d x} = y ({sec}^{2} x ln cos x - {tan}^{2} x)$ $dy/dx=y(sec^2xlncosx - tan^2x)$
$= ({sec}^{2} x ln cos x - {tan}^{2} x) {cos x}^{tan x}$ $=(sec^2xlncosx-tan^2x)cosx^tanx$

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How do you find the derivative of ${cos x}^{tan x}$ $cosx^tanx$ ?

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