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: Derivative Evaluation – differentiate $f(x) = x \tan^{-1} x$ using the product rule, then substitute $x = 1$.

Step 1: Apply product rule:

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

Step 2: Simplify:

$f'(x) = \tan^{-1} x + \frac{x}{1+x^2}$.

Step 3: Substitute $x = 1$: …

The key here is recognizing that $f(x)$ is a product of two functions: $x$ and $\tan^{-1} x$ (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 $\tan^{-1} x$ is a standard result: $\frac{d}{dx} \tan^{-1} x = \frac{1}{1+x^2}$. 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 $v(x) = \tan^{-1} x$.

   Then $u'(x) = 1$, and $v'(x) = \frac{1}{1+x^2}$.

   So:

2. Evaluate at $x = 1$. Substitute $x = 1$ into the derivative:

Now, $\tan^{-1}(1)$ is the angle whose tangent is 1. That angle is $\frac{\pi}{4}$ (since $\tan \frac{\pi}{4} = 1$).

Therefore:

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). …

Method: Differentiating a Product Involving an Inverse Trigonometric Function

Use this method whenever $f(x)$ is a product of a simple algebraic factor (like $x$) and an inverse trig function (like $\tan^{-1}x$, $\sin^{-1}x$, etc.), and you need $f'(a)$ at a specific point.

Steps

Step 1: Recognize the product structure

Identify the two factors being multiplied, $u(x)$ and $v(x)$, so you know to reach for the product rule rather than trying to simplify the expression first (there is usually nothing to simplify before differentiating).

Step 2: Recall the standard derivative of the inverse trig factor

Memorize (or quickly re-derive) the standard results:

$\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}, \quad \frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1-x^2}}, \quad \frac{d}{dx}\cos^{-1}x = \frac{-1}{\sqrt{1-x^2}}.$ …

Common Mistakes

Mistake 1: Skipping the product rule

A student might try to differentiate $x\tan^{-1}x$ as if it were a single function, or differentiate only the $\tan^{-1}x$ part and ignore the leading $x$. Why it's wrong: whenever two functions of $x$ are multiplied together, both factors contribute to the derivative through the product rule; dropping one term silently loses information. Correct approach: explicitly label $u(x)=x$ and $v(x)=\tan^{-1}x$ before differentiating, so both terms are accounted for.

Mistake 2: Confusing the derivative of $\tan^{-1}x$ with the derivative of $\tan x$ …