If u=cos ^3 x and v=sin ^3 x,then left(frac{d | vedclass

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(D) Given $u = \cos^3 x$ and $v = \sin^3 x$. First,we find the derivatives with respect to $x$: $\frac{du}{dx} = 3 \cos^2 x \cdot (-\sin x) = -3 \cos^

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-$1$

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(D) Given $u = \cos^3 x$ and $v = \sin^3 x$. First,we find the derivatives with respect to $x$: $\frac{du}{dx} = 3 \cos^2 x \cdot (-\sin x) = -3 \cos^2 x \sin x$ $\frac{dv}{dx} = 3 \sin^2 x \cdot (\cos x) = 3 \sin^2 x \cos x$ Now,using the chain rule for parametric differentiation: $\frac{dv}{du} = \frac{dv/dx}{du/dx} = \frac{3 \sin^2 x \cos x}{-3 \cos^2 x \sin x} = -\frac{\sin x}{\cos x} = -	an x$ Evaluating at $x = \frac{\pi}{4}$: $\left(\frac{dv}{du}ight)_{x=\frac{\pi}{4}} = -	an\left(\frac{\pi}{4}ight) = -1$

If $u=\cos ^3 x$ and $v=\sin ^3 x$,then $\left(\frac{d v}{d u}\right)_{x=\frac{\pi}{4}}$ is equal to

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