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Question 83 of 99

Q.If f(x)=xx+1f(x)=\dfrac{x}{x+1}, then its differential is :

(a) 1x+1dx\dfrac{1}{x+1}dx
(b) −1(x+1)2dx\dfrac{-1}{(x+1)^2}dx
(c) −1x+1dx\dfrac{-1}{x+1}dx
(d) 1(x+1)2dx\dfrac{1}{(x+1)^2}dx
Puducherry TnboardTamil Nadu HSC (DGE) Board 2022MCQ· 1mImportance★★★★★
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Differentiating f(x)=xx+1f(x)=\dfrac{x}{x+1} by the quotient rule gives f′(x)=1(x+1)2f'(x)=\dfrac{1}{(x+1)^2}, so dy=1(x+1)2dxdy=\dfrac{1}{(x+1)^2}dx.

  1. Let y=f(x)=xx+1y=f(x)=\dfrac{x}{x+1}.
  2. By the quotient rule, dydx=(x+1)⋅ddx(x)−x⋅ddx(x+1)(x+1)2\dfrac{dy}{dx}=\dfrac{(x+1)\cdot\dfrac{d}{dx}(x)-x\cdot\dfrac{d}{dx}(x+1)}{(x+1)^2}. …

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