One maximum point of {sin ^p}x{cos ^q}x is | vedclass

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(A) Let $y = {\sin ^p}x{\cos ^q}x$. Taking the derivative with respect to $x$: $\frac{dy}{dx} = p{\sin ^{p - 1}}x(\cos x){\cos ^q}x + {\sin ^p}x(q{\co

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

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(A) Let $y = {\sin ^p}x{\cos ^q}x$. Taking the derivative with respect to $x$: $\frac{dy}{dx} = p{\sin ^{p - 1}}x(\cos x){\cos ^q}x + {\sin ^p}x(q{\cos ^{q - 1}}x)(-\sin x)$ $\frac{dy}{dx} = p{\sin ^{p - 1}}x{\cos ^{q + 1}}x - q{\cos ^{q - 1}}x{\sin ^{p + 1}}x$ For critical points,set $\frac{dy}{dx} = 0$: $p{\sin ^{p - 1}}x{\cos ^{q + 1}}x = q{\cos ^{q - 1}}x{\sin ^{p + 1}}x$ Divide both sides by ${\sin ^{p - 1}}x{\cos ^{q - 1}}x$: $p{\cos ^2}x = q{\sin ^2}x$ $\frac{{\sin ^2}x}{{\cos ^2}x} = \frac{p}{q}$ ${	an ^2}x = \frac{p}{q}$ $	an x = \sqrt{\frac{p}{q}}$ $x = {	an ^{ - 1}}\sqrt{\frac{p}{q}}$

One maximum point of ${\sin ^p}x{\cos ^q}x$ is

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