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Question 281 of 281

Q.If y=sin⁡(sin⁡x)y = \sin(\sin x), prove that d2ydx2+tan⁡x dydx+ycos⁡2x=0\dfrac{d^2 y}{dx^2} + \tan x\, \dfrac{dy}{dx} + y\cos^2 x = 0.

Sikkim CbseCBSE Class XII Board 2018Subjective· 4mImportance★★★★★
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Substituting y′y' and y′′y'' makes every term cancel, proving the identity.

Concept. Repeated use of the chain and product rules.

Why this method. Compute the two derivatives explicitly and plug into the left side.

Working. y=sin⁡(sin⁡x)y=\sin(\sin x).

dydx=cos⁡(sin⁡x)⋅cos⁡x.\frac{dy}{dx}=\cos(\sin x)\cdot\cos x.

d2ydx2=−sin⁡(sin⁡x)cos⁡x⋅cos⁡x+cos⁡(sin⁡x)⋅(−sin⁡x)=−sin⁡(sin⁡x)cos⁡2x−sin⁡xcos⁡(sin⁡x).\frac{d^2y}{dx^2}=-\sin(\sin x)\cos x\cdot\cos x+\cos(\sin x)\cdot(-\sin x)=-\sin(\sin x)\cos^2x-\sin x\cos(\sin x).

Now

tan⁡x dydx=tan⁡x⋅cos⁡(sin⁡x)cos⁡x=sin⁡xcos⁡(sin⁡x),\tan x\,\frac{dy}{dx}=\tan x\cdot\cos(\sin x)\cos x=\sin x\cos(\sin x), …

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