The function f(x) = sin^p x cos^q x has a max | vedclass

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(A) Let $y = \sin^p x \cos^q x$.Taking the natural logarithm on both sides,we get $z = \ln(y) = p \ln(\sin x) + q \ln(\cos x)$.Differentiating with re

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$x = 	an^{-1} \sqrt{p/q}$

Vedclass · Answer

(A) Let $y = \sin^p x \cos^q x$.Taking the natural logarithm on both sides,we get $z = \ln(y) = p \ln(\sin x) + q \ln(\cos x)$.Differentiating with respect to $x$,we have $\frac{dz}{dx} = p \cot x - q 	an x$.For critical points,set $\frac{dz}{dx} = 0$,which gives $p \cot x = q 	an x$,or $	an^2 x = p/q$. Thus,$x = 	an^{-1} \sqrt{p/q}$.To check for a maximum,we find the second derivative: $\frac{d^2z}{dx^2} = -p \csc^2 x - q \sec^2 x$.Since $p, q > 0$,$\frac{d^2z}{dx^2} < 0$ for all $x$ in the domain,confirming that $x = 	an^{-1} \sqrt{p/q}$ is a point of local maximum.

The function $f(x) = \sin^p x \cos^q x$ has a maximum at:

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