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

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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}]#