Question

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

Answer 1

Given, $y=\sec x+\tan x$

$\frac{dy}{dx}=\sec x\times \tan x+{\sec }^{2}x$

$\frac{d^{2}y}{d{x}^2}=\sec x\times \left({\sec }^{2}x\right)+\tan x\times (\sec x\times \tan x)+2\sec x\times \sec x\times \tan x$

$\frac{d^{2}y}{d{x}^2}={\sec }^{3}x+\sec x{\tan }^{2}x+2{\sec }^{2}x\tan x$

$\frac{d^{2}y}{d{x}^2}=\sec x({\sec }^{2}x+{\tan }^{2}x+2\sec x\tan x)$

$\frac{d^{2}y}{d{x}^2}=\sec x(\sec x+\tan x{)}^2$

Putting $\sec x=\frac{1}{\cos x}$ and $\tan x=\frac{\sin x}{\cos x}$ and solving,

$\frac{d^{2}y}{d{x}^2}=\frac{1}{\cos x}{(\frac{1}{\cos x}+\frac{\sin x}{\cos x})}^2$

$=\frac{(1+\sin x{)}^2}{\cos x\times {\cos }^{2}x}$

$=\frac{(1+\sin x{)}^2}{\cos x\times (1-{\sin }^{2}x)}$

$=\frac{1}{\cos x}\left[\frac{1+\sin x}{1-\sin x}\right]$

Multiply and divide by $(1-\sin x)$, we get

$\frac{1}{\cos x}\left[\frac{1-{\sin }^{2}x}{(1-\sin x{)}^2}\right]$

$=\frac{\cos x}{(1-\sin x{)}^2}$