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 $