Question:
If \( y = \frac{x}{x+1} + \frac{x+1}{x} \), then \( \frac{d^2y}{dx^2} \) at \( x=1 \) is equal to
If \( y = \frac{x}{x+1} + \frac{x+1}{x} \), then \( \frac{d^2y}{dx^2} \) at \( x=1 \) is equal to
Show Hint
Always simplify rational expressions before differentiating.
Updated On: May 8, 2026
- \( \frac{7}{4} \)
- \( \frac{7}{8} \)
- \( \frac{1}{4} \)
- \( -\frac{7}{8} \)
- \( -\frac{7}{4} \)
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The Correct Option is B
Solution and Explanation
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: \[ \boxed{\frac{7}{4}} \]
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: \[ \boxed{\frac{7}{4}} \]
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