If 𝑓(𝑥) = 𝑥 tan−1 𝑥 , then 𝑓′(1)is equal to
(A) 𝜋 4 − 1 2
(B) 𝜋 4 + 1 2
(C) − 𝜋 4 − 1 2
(D) − 𝜋 4 + 1 2

Question

Accepted Answer

> [!TLDR]
> The derivative of $f(x) = x 	an^{-1} x$ is found using the product rule. Evaluating at $x=1$ gives $f'(1) = \frac{\pi}{4} + \frac{1}{2}$, which corresponds to option **(B)**.

The key here is recognizing that $f(x)$ is a product of two functions: $x$ and $	an^{-1} x$ (inverse tangent, also written as $\arctan x$). When you see a product, your first instinct should be the product rule — not expanding or simplifying, because there’s nothing to simplify here. The derivative of $	an^{-1} x$ is a standard result: $\frac{d}{dx} 	an^{-1} x = \frac{1}{1+x^2}$. That’s the only “tricky” part; everything else is straightforward algebra.

Let’s walk through it.

1. **Apply the product rule.**  
   For $f(x) = u(x) \cdot v(x)$, we have $f'(x) = u'(x) v(x) + u(x) v'(x)$.  
   Here, let $u(x) = x$ and $v(x) = 	an^{-1} x$.  
   Then $u'(x) = 1$, and $v'(x) = \frac{1}{1+x^2}$.

So:
   $$f'(x) = (1) \cdot 	an^{-1} x + x \cdot \frac{1}{1+x^2} = 	an^{-1} x + \frac{x}{1+x^2}.$$

2. **Evaluate at $x = 1$.**  
   Substitute $x = 1$ into the derivative:
   $$f'(1) = 	an^{-1}(1) + \frac{1}{1+1^2} = 	an^{-1}(1) + \frac{1}{2}.$$

Now, $	an^{-1}(1)$ is the angle whose tangent is 1. That angle is $\frac{\pi}{4}$ (since $	an \frac{\pi}{4} = 1$).  
   Therefore:
   $$f'(1) = \frac{\pi}{4} + \frac{1}{2}.$$

3. **Match with the options.**  
   The options are given as combinations of $\frac{\pi}{4}$ and $\frac{1}{2}$ with plus/minus signs. Our result $\frac{\pi}{4} + \frac{1}{2}$ exactly matches option **(B)**. …

Q.If 𝑓(𝑥) = 𝑥 tan−1 𝑥 , then 𝑓′(1)is equal to
(A) 𝜋 4 − 1 2
(B) 𝜋 4 + 1 2
(C) − 𝜋 4 − 1 2
(D) − 𝜋 4 + 1 2

Concept understanding — Derivative Evaluation

Derivative Evaluation

The geometric picture

The limit definition

A worked evaluation

From a number to a function …

Q.If 𝑓(𝑥) = 𝑥 tan−1 𝑥 , then 𝑓′(1)is equal to (A) 𝜋 4 − 1 2 (B) 𝜋 4 + 1 2 (C) − 𝜋 4 − 1 2 (D) − 𝜋 4 + 1 2