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

Question

Charles Arocha · Accepted Answer

#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}]#

Emma Aldridge · Answer

undefined To find the derivative of $(\cos x)^{\sin x}$, we use the chain rule. Let $ u = \cos x $ and $ v = \sin x $. Then the function becomes $ u^v $. The derivative is given by $ \frac{d}{dx}(u^v) = vu^{v-1}\frac{du}{dx} + u^v \ln(u)\frac{dv}{dx} $. Substituting back $ u = \cos x $ and $ v = \sin x $, we get the derivative as $ (\sin x)(\cos x)^{\sin x - 1}(-\sin x) + (\cos x)^{\sin x}\ln(\cos x)\cos x $. Simplify this expression to get the final derivative.