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

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  • $y=f(x)$
  • $\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.
answered Sep 24, 2013 by sreemathi.v

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