Question
$f(x) = \frac{1}{x+1}, x \neq -1$
✓
Answer
Here, $f(x) = \frac{1}{x+1}$
$\therefore f(x + h) = f(x)$
Hence, $f(x) = \frac{1}{x+1}$ then $f'(x) = -\frac{1}{(x+1)^{2}}$.
$\therefore f(x + h) = f(x)$
Hence, $f(x) = \frac{1}{x+1}$ then $f'(x) = -\frac{1}{(x+1)^{2}}$.
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