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Question

Let y=(3x−13x+1)sinx+loge(1+x),x>−1. Then, at x=0,dydx equals

A
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Solution

The correct option is A 1
Given,
y=(3x13x+1)sinx+loge(1+x),x>1

Differentiating w.r.t. x, we get
dydx=ddx[(3x13x+1)sinx]+ddxloge(1+x)

=(3x13x+1)ddxsinx+sinxddx(3x13x+1)+11+xddx(1+x)

=(3x13x+1)cosx+sinx (3x+1)ddx(3x1)(3x1)ddx(3x+1)(3x+1)2+11+x

=(3x13x+1)cosx+sinx(3x+1)(3xloge3)(3x1)(3xloge3)(3x+1)2+11+x

(dydx)at x=0=0+0+11+0=1

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