Question:
If
\[
y = \frac{x}{x+1} + \frac{x+1}{x},
\]
then
\[
\frac{d^2 y}{dx^2} \text{ at } x = 1 \text{ is equal to:}
\]
If
\[
y = \frac{x}{x+1} + \frac{x+1}{x},
\]
then
\[
\frac{d^2 y}{dx^2} \text{ at } x = 1 \text{ is equal to:}
\]
Show Hint
Simplify the function before differentiating to reduce errors.
Updated On: Mar 23, 2026
- \( \dfrac{7}{4} \)
- \( \dfrac{7}{8} \)
- \( \dfrac{1}{4} \)
- -(7)/(8)
Show Solution
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The Correct Option is D
Solution and Explanation
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: Differentiate:
\[ \frac{dy}{dx} = -\frac{1}{x^2} + \frac{1}{(x+1)^2} \]
Step 3: Second derivative:
\[ \frac{d^2 y}{dx^2} = \frac{2}{x^3} - \frac{2}{(x+1)^3} \]
Step 4: At \(x = 1\):
\[ \frac{d^2 y}{dx^2} = 2 - \frac{2}{8} = \frac{14}{8} - \frac{2}{8} = \frac{12}{8} = \frac{3}{2} \]
\[ y = \frac{x}{x+1} + \frac{x+1}{x} = 1 - \frac{1}{x+1} + 1 + \frac{1}{x} = 2 + \frac{1}{x} - \frac{1}{x+1} \]
Step 2: Differentiate:
\[ \frac{dy}{dx} = -\frac{1}{x^2} + \frac{1}{(x+1)^2} \]
Step 3: Second derivative:
\[ \frac{d^2 y}{dx^2} = \frac{2}{x^3} - \frac{2}{(x+1)^3} \]
Step 4: At \(x = 1\):
\[ \frac{d^2 y}{dx^2} = 2 - \frac{2}{8} = \frac{14}{8} - \frac{2}{8} = \frac{12}{8} = \frac{3}{2} \]
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