Question

If $y=\sin x$ , then $\frac{d^2}{d{y}^2}({\cos }^{7}x)$ is equal to

• A. $7{\cos }^{5}x+35{\cos }^{3}x$
• B. $7{\cos }^{5}x+35{\cos }^{2}x$
• C. $-7{\cos }^{5}x+35{\cos }^{2}x$
• D. None of the above

Answer 1

Answer: (D)

We have, $y=\sin x$
$\Rightarrow$ $\frac{dy}{dx}=\cos x$
Now, $\frac{d^2}{d{y}^2}({\cos }^{7}x)$
$=\frac{d}{dy}\left[\frac{d}{dy}({\cos }^{7}x)\right]$
$=\frac{d}{dy}\left[7{\cos }^{6}x(-\sin x)\frac{dx}{dy}\right]$
$=\frac{d}{dy}\left[-7{\cos }^{6}x\sin x\cdot \frac{1}{\cos x}\right]\left(\because \frac{dx}{dy}=\frac{1}{\cos x}\right)$
$=\frac{d}{dy}[-7{\cos }^{5}x\sin x]$
$=-7[-5{\cos }^{4}x\cdot \sin x\cdot \sin x+{\cos }^{5}x\cdot \cos x]\frac{dx}{dy}$
$=-7[-5{\cos }^{3}x{\sin }^{2}x+{\cos }^{5}x]\left(\because \frac{dx}{dy}=\frac{1}{\cos x}\right)$
$=35{\cos }^{3}x{\sin }^{2}x-7{\cos }^{5}x$
$=-7{\cos }^{5}x+35{\cos }^{3}x(1-{\cos }^{2}x)$
$=-42{\cos }^{5}x+35{\cos }^{3}x$