How do you find the derivative of $(cos x)^(sin x)$ ?

Question

1456 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-13T15:25:03

$y' = (cos x)^{sin x} cdot [cos x cdot ln( cos x) + cos x - sec x]$

$a^b = exp ln a^b = exp (b ln a)$

$y = exp(sinx cdot ln( cos x))$

$frac{dy}{dx} = exp(sinx cdot ln( cos x)) cdot \frac{d}{dx} (sinx cdot ln( cos x))$

$= (cos x)^{sin x} cdot [\frac{d}{dx} (sinx) cdot ln( cos x) + sin x cdot frac{d}{dx} (ln (cos x))]$

$= (cos x)^{sin x} cdot [cos x cdot ln( cos x) + sin x cdot \frac{1}{cos x} cdot frac{d}{dx} (cos x)]$

$= (cos x)^{sin x} cdot [cos x cdot ln( cos x) - \frac{sin^2 x}{cos x}]$

$= (cos x)^{sin x} cdot [cos x cdot ln( cos x) + \frac{cos^2 x - 1}{cos x}]$

How do you find the derivative of (cos x)^(sin x)(cos x)^(sin x)?