Question

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

• A. $-1/2$
• B. $3/2$
• C. $5/2$
• D. $-3/2$

Answer 1

Answer: (C)

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