Differentiate sin ^2 x with respect to cos ^2 | vedclass

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(C) Let $u = \sin ^2 x$ and $v = \cos ^2 x$. We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx}$. First,differentiate $u$ with respect to $x$: $\fra

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(C) Let $u = \sin ^2 x$ and $v = \cos ^2 x$. We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx}$. First,differentiate $u$ with respect to $x$: $\frac{du}{dx} = \frac{d}{dx}(\sin ^2 x) = 2 \sin x \cos x = \sin(2x)$. Next,differentiate $v$ with respect to $x$: $\frac{dv}{dx} = \frac{d}{dx}(\cos ^2 x) = 2 \cos x (-\sin x) = -2 \sin x \cos x = -\sin(2x)$. Now,calculate the derivative: $\frac{du}{dv} = \frac{\sin(2x)}{-\sin(2x)} = -1$.

Differentiate $\sin ^2 x$ with respect to $\cos ^2 x$.

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