The derivative of sin^2 x with respect to cos | vedclass

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Question

(D) Let $y = \sin^2 x$ and $z = \cos^2 x$. First,differentiate $y$ with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(\sin^2 x) = 2 \sin x \cos x = \s

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(D) Let $y = \sin^2 x$ and $z = \cos^2 x$. First,differentiate $y$ with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(\sin^2 x) = 2 \sin x \cos x = \sin 2x$. Next,differentiate $z$ with respect to $x$: $\frac{dz}{dx} = \frac{d}{dx}(\cos^2 x) = 2 \cos x (-\sin x) = -2 \sin x \cos x = -\sin 2x$. Now,find the derivative of $y$ with respect to $z$ using the chain rule: $\frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{\sin 2x}{-\sin 2x} = -1$. Thus,the correct option is $D$.

The derivative of $\sin^2 x$ with respect to $\cos^2 x$ is

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