Question

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

Answer 1

$z=(\cos x{)}^5\Rightarrow \frac{dz}{dx}=-5(\cos x{)}^4\sin x$
$y=\sin x\Rightarrow \frac{dy}{dx}=\cos x$
So,
$\frac{dz}{dy}=-5{\cos }^{3}x\sin x\Rightarrow \frac{d^{2}z}{d{y}^2}=\frac{d}{dx}\left(\frac{dz}{dy}\right)\cdot \frac{dx}{dy}\Rightarrow \frac{d^{2}z}{d{y}^2}=\frac{d}{dx}[-5{\cos }^{3}x\sin x]\cdot \frac{1}{\cos x}\Rightarrow \frac{d^{2}z}{d{y}^2}=\frac{-5}{\cos x}[{\cos }^{4}x-3{\cos }^{2}x{\sin }^{2}x]\Rightarrow \frac{d^{2}z}{d{y}^2}=-5[{\cos }^{3}x-3\cos x(1-{\cos }^{2}x\left)\right]\Rightarrow \frac{d^{2}z}{d{y}^2}=-5[4{\cos }^{3}x-3\cos x]\Rightarrow \frac{d^{2}z}{d{y}^2}=-5\cos 3x$
Putting $x=\frac{2\pi }{9}$
$\frac{d^{2}z}{d{y}^2}=-5\cos \frac{3\times 2\pi }{9}=\frac{5}{2}$