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  4. Let z=( cos x)5 and y= sin x. Then the value of (d2 z/d y2) at x=(2 π/9) is

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Q. Let $z=(\cos x)^5$ and $y=\sin x$. Then the value of $\frac{d^2 z}{d y^2}$ at $x=\frac{2 \pi}{9}$ is

Continuity and Differentiability

Solution:

$z=(\cos x)^5 ; y=\sin x$ $\frac{ dz }{ dx }=-5 \cos ^4 x \cdot \sin x ; \frac{ dy }{ dx }=\cos x $ $\therefore \frac{ dz }{ dy }=-5 \cos ^3 x \cdot \sin x $ $\text { now } \frac{ d ^2 z }{ dy ^2}=\frac{ d }{ dx }\left(\frac{ dz }{ dy }\right) \cdot \frac{ dx }{ dy }$ $ =-5 \frac{d}{d x}\left[\cos ^3 x \cdot \sin x\right] \frac{1}{\cos x}=-5\left[\cos ^4 x-3 \sin ^2 x \cdot \cos ^2 x\right] \frac{1}{\cos x}$ $ =-5\left(\cos ^3 x-3 \sin ^2 x \cdot \cos x\right)=-5\left(\cos ^3 x-3 \cos x\left(1-\cos ^2 x\right)\right)$ $ =-5\left(4 \cos ^3 x-3 \cos x\right)=-5 \cos 3 x$ $\therefore \left.\frac{ d ^2 z}{ dy ^2}\right|_{x=\frac{2 \pi}{9}}=-5 \cos 120^{\circ}=\frac{5}{2}$