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

Question

If $t = e^{2x}$ and $y = \ln(t^2)$, then $\frac{d^2 y}{dx^2}$ is:

Answer 1

Understanding the Concept: Differentiate twice.

Step 1: Simplify
\[ y = \frac{x}{x+1} + \frac{x+1}{x} \] \[ = 1 - \frac{1}{x+1} + 1 + \frac{1}{x} \] \[ = 2 + \frac{1}{x} - \frac{1}{x+1} \]

Step 2: First derivative
\[ y' = -\frac{1}{x^2} + \frac{1}{(x+1)^2} \]

Step 3: Second derivative
\[ y'' = \frac{2}{x^3} - \frac{2}{(x+1)^3} \]

Step 4: Substitute $x=1$
\[ y'' = 2 - \frac{2}{8} = 2 - \frac{1}{4} = \frac{7}{4} \] Thus: \[ \boxed{\frac{7}{4}} \]

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

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The Correct Option is B

Solution and Explanation

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