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Question

Mathematics Question on Differentiability

If y=xx+1+x+1x,y=\frac{x}{x+1}+\frac{x+1}{x}, then d2ydx2\frac{d^{2}y}{dx^{2}} at x=1x=1 is equal to

A

74\frac{7}{4}

B

78\frac{7}{8}

C

14\frac{1}{4}

D

−78\frac{-7}{8}

Answer

74\frac{7}{4}

Explanation

Solution

y=xx+1+x+1xy=\frac{x}{x+1}+\frac{x+1}{x}
⇒dydx=(1)(x+1)−x(1)(x+1)2\Rightarrow \frac{d y}{d x}=\frac{(1)(x+1)-x(1)}{(x+1)^{2}}
+(1)(x)−(x+1)⋅1x2+\frac{(1)(x)-(x+1) \cdot 1}{x^{2}}
=1(x+1)2−1x2=\frac{1}{(x+1)^{2}}-\frac{1}{x^{2}}
⇒d2ydx2=−2(1+x)3+2x3\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{-2}{(1+x)^{3}}+\frac{2}{x^{3}}
On putting x=1x=1, we get
(d2ydx2)x=1=−2(1+1)3+2(1)3\left(\frac{d^{2} y}{d x^{2}}\right)_{x=1} =\frac{-2}{(1+1)^{3}}+\frac{2}{(1)^{3}}
=−28+2=\frac{-2}{8}+2
=−14+2=74=-\frac{1}{4}+2=\frac{7}{4}