Limits, Continuity & Differentiability
Differentiation
nta_abhyas_2025
Grade 12
Question:
If $f(x) = \sqrt{x - 4\sqrt{x - 4}} + \tan^{-1}\left(\frac{1+x}{1-x}\right), 4 < x < 8$, then the value of $f'(5)$ is equal to
$-\frac{1}{13}$
$0$
$\frac{1}{13}$
$-\frac{1}{13}$
Step-by-Step Solution
Key Concept: Differentiation of composite functions requires careful application of chain rule and simplification of nested functions
Simplify $f(x) = \sqrt{(x-3)^2} - x^2 - 2\sqrt{x} - 4 + \tan^{-1}\left(\frac{2x}{1+x}\right)$. For $x > 3$, $\sqrt{(x-3)^2} = x - 3$. So $f(x) = x - 3 - x^2 - 2\sqrt{x} - 4 + \tan^{-1}\left(\frac{2x}{1+x}\right)$. The derivative is: $f'(x) = 1 - 2x - \frac{1}{\sqrt{x}} + \frac{2(1+x) - 2x}{(1+x)^2 + 4x^2} = 1 - 2x - \frac{1}{\sqrt{x}} + \frac{2}{1 + 2x + 2x^2}$. At $x = 5$: $f'(5) = 1 - 10 - \frac{1}{\sqrt{5}} + \frac{2}{1 + 10 + 50} = -9 - \frac{1}{\sqrt{5}} + \frac{2}{61} = -\frac{7}{6}$.
Correct Answer: -7/6