Browse Questions

# If $y=e^{\large \tan\: x},$ prove that $\cos^2x\large\frac{d^2y}{dx}$$-(1+\sin\: 2x)\large\frac{dy}{dx}$$=0$

Toolbox:
• $y=f(x)$
• $\large\frac{dy}{dx}$$=f'(x) • \large\frac{d^2y}{dx^2}=\frac{d}{dx}\big(\frac{dy}{dx}\big) • \sin 2x=2\sin x\cos x Step 1: Given : y=e^{\tan x} Differentiating with respect to y we get, \large\frac{dy}{dx}$$=e^{\tan x}.\sec^2x$
(i.e)$\large\frac{1}{\sec^2x}\frac{dy}{dx}=e^{\tan x}$
$\cos^2 x\large\frac{dy}{dx}=$$y Step 2: Differentiating again with respect to x we get, Apply product rule \cos^2 x\large\frac{d^2y}{dx^2}+\frac{dy}{dx}$$(2\cos x(-\sin x))=\large\frac{dy}{dx}$
$\cos^2 \large\frac{d^2y}{dx^2}+\frac{dy}{dx}$$(-2\sin x\cos x-1)=0 But \sin 2x=2\sin x\cos x \Rightarrow \cos^2\large\frac{d^2y}{dx^2}-\frac{dy}{dx}$$(1+\sin 2x)=0$
Hence proved.