Question

If $y = \frac{x}{x+1} + \frac{x+1}{x} , $ then $\frac{d^{2}y}{dx^{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}{dx^{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}{dx^{2}}\bigg|_{x=1} = \frac{-2}{\left(1+1\right)^{3}} + \frac{2}{\left(1\right)^{3}} = \frac{-2}{8} +2 = \frac{7}{4} $