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 · 2017-02-19T20:01:56

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

Explanation:

Write the function as:

${(cos x)}^{tan x} = {(e^{ln (cos x)})}^{tan x} = e^{ln (cos x) tan x}$ $(cosx)^(tanx) = (e^(ln(cosx)))^(tanx) = e^(ln(cosx)tanx)$

Now differentiate using the chain rule:

$\frac{d}{d x} (e^{ln (cos x) tan x}) = e^{ln (cos x) tan x} \frac{d}{d x} (ln (cos x) tan x)$ $d/dx (e^(ln(cosx)tanx)) = e^(ln(cosx)tanx) d/dx(ln(cosx)tanx)$

$\frac{d}{d x} (e^{ln (cos x) tan x}) = e^{ln (cos x) tan x} (tan x \frac{d}{d x} (ln (cos x)) + ln (cos x) \frac{d}{d x} (tan x))$ $d/dx (e^(ln(cosx)tanx)) = e^(ln(cosx)tanx) (tanx d/dx(ln(cosx))+ ln(cosx)d/dx(tanx))$

$\frac{d}{d x} (e^{ln (cos x) tan x}) = e^{ln (cos x) tan x} (tan x (- \frac{sin x}{cos x}) + \frac{ln (cos x)}{{cos}^{2} x})$ $d/dx (e^(ln(cosx)tanx)) = e^(ln(cosx)tanx) (tanx (-sinx/cosx)+ ln(cosx)/cos^2x)$

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

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

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