Question

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

• A. $\frac{7}{4}$
• B. $\frac{7}{8}$
• C. $\frac{1}{4}$
• D. $\frac{-7}{8}$

Answer 1

Answer: (A)

Given expression can be written as
$y=1-\frac{1}{x+1}+1+\frac{1}{x}\Rightarrow \frac{dy}{dx}=\frac{1}{\left(1+x^2\right)}-\frac{1}{x^2}$
$\frac{d^{2}y}{d{x}^2}=-2{\left(1+x\right)}^{-3}+2x^{-3}=\frac{-2}{{\left(1+x\right)}^3}+\frac{2}{x^3}$
Now, $\frac{d^{2}y}{d{x}^2}{|}_{x=1}=\frac{-2}{{\left(1+1\right)}^3}+\frac{2}{{\left(1\right)}^3}=\frac{-2}{8}+2=\frac{7}{4}$