Question

If $y=\tan x+\sec x$, then prove that $\frac{d^{2}y}{d{x}^2}=\frac{\cos x}{{(1-\sin x)}^2}$

Answer 1

Given: $y=\tan x+\sec x$
Prove that: $\frac{d^{2}y}{d{x}^2}=\frac{\cos x}{{(1-\sin x)}^2}$
$y=\frac{\sin x}{\cos x}+\frac{1}{\cos x}$
$=\frac{1+\sin x}{\cos x}$
differentiate with respect to $x$
$\frac{dy}{dx}=\frac{d}{dx}\left(\frac{1+\sin x}{\cos x}\right)$
$\frac{dy}{dx}=\frac{\cos x\frac{d}{dx}(1+\sin x)-(1+\sin x)\frac{d}{dx}\cos x}{{\cos }^{2}x}$
$=\frac{{\cos }^{2}x+\sin x+{\sin }^{2}x}{{\cos }^{2}x}=\frac{1+\sin x}{{\cos }^{2}x}=\frac{1+\sin x}{1-{\sin }^{2}x}$
$=\frac{1+\sin x}{(1+\sin x)(1-\sin x)}=\frac{1}{1-\sin x}$
differentiate with respect to $x$
$\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(\frac{1}{1-\sin x}\right)$
$\frac{d^{2}y}{d{x}^2}=\frac{(1-\sin x)\frac{d}{dx}\left(1\right)-\left(1\right)\frac{d}{dx}(1-\sin x)}{{(1-\sin x)}^2}$
$=\frac{\cos x}{{(1-\sin x)}^2}$
Hence proved.