If x=sin theta and y=sin^3 theta,then the val | vedclass

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(D) Given that $x = \sin 	heta$ and $y = \sin^3 	heta$. Since $y = (\sin 	heta)^3$,we can substitute $x = \sin 	heta$ to get $y = x^3$. Now,differ

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$6$

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(D) Given that $x = \sin 	heta$ and $y = \sin^3 	heta$. Since $y = (\sin 	heta)^3$,we can substitute $x = \sin 	heta$ to get $y = x^3$. Now,differentiate $y$ with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(x^3) = 3x^2$. Next,find the second derivative $\frac{d^2 y}{dx^2}$: $\frac{d^2 y}{dx^2} = \frac{d}{dx}(3x^2) = 6x$. At $	heta = \frac{\pi}{2}$,the value of $x$ is $x = \sin(\frac{\pi}{2}) = 1$. Substituting $x = 1$ into the second derivative: $\left(\frac{d^2 y}{dx^2}ight)_{x=1} = 6(1) = 6$.

If $x=\sin \theta$ and $y=\sin^3 \theta$,then the value of $\frac{d^2 y}{dx^2}$ at $\theta=\frac{\pi}{2}$ is:

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