Find the derivative of $\tan^{-1} x$ with respect to $\log x$; (where $x \in (1, \infty)$).

> [!TLDR] > We want the derivative of $\tan^{-1} x$ **with respect to** $\log x$, not with respect to $x$. Using the chain rule in parametric form, the answer is $\frac{x}{1+x^2}$. The key idea here is that "derivative with respect to $\log x$" is not the same as the ordinary derivative $\frac{d}{dx}$. When we say "differentiate $u$ with respect to $v$", we mean $\frac{du}{dv}$. So we need $\frac{d(\tan^{-1} x)}{d(\log x)}$. This is a classic case of parametric differentiation: if both $u$ and $v$ are functions of $x$, then $$ \frac{du}{dv} = \frac{du/dx}{dv/dx}. $$ Let’s apply that. 1. Let $u = \tan^{-1} x$ and $v = \log x$. We want $\frac{du}{dv}$. 2. First, find $\frac{du}{dx}$: $$ \frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}. $$ This is a standard result — the derivative of the inverse tangent. 3. Next, find $\frac{dv}{dx}$: $$ \frac{d}{dx}(\log x) = \frac{1}{x}. $$ Since $x > 1$, $\log x$ is the natural logarithm (base $e$), and its derivative is $1/x$. 4. Now, using the parametric formula: $$ \frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{\frac{1}{1+x^2}}{\frac{1}{x}} = \frac{1}{1+x^2} \cdot \frac{x}{1} = \frac{x}{1+x^2}. $$ > [!WARNING] > A common mistake is to compute $\frac{d}{dx}(\tan^{-1} x)$ and stop, forgetting that the question asks for the derivative **with respect to $\log x$**, not $x$. Always check the "denominator" of the derivative. > [!TIP] > If you ever see "derivative of $f(x)$ with respect to $g(x)$", immediately think $\frac{f'(x)}{g'(x)}$ — provided $g'(x) \neq 0$. This works because both are functions of the same underlying variable $x$. > [!ANSWER] > The derivative of $\tan^{-1} x$ with respect to $\log x$ is $\boxed{\frac{x}{1+x^2}}$.

Find the derivative of tan⁻¹ x with respect to log x; (where x ∈ (1, ∞)).

Question

\tan^{-1} x

CBSESample paperShort· 2mImportance★★★★★

✓ Free question

Concept understanding — Inverse Tangent Identity

Inverse Tangent Identities

The inverse tangent function obeys a family of addition and doubling identities that let you combine two arctangents into one. They come straight from the tangent addition formula, but they carry conditions you must respect.

The core addition identity

Start from $\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ . Put $A=\tan^{-1}x$ and $B=\tan^{-1}y$ , so $\tan A=x$ and $\tan B=y$ . Then

\tan^{-1}x + \tan^{-1}y = \tan^{-1}\!\left(\frac{x+y}{1-xy}\right), \qquad xy<1.

The restriction $xy<1$ keeps the combined angle inside the principal range $(-\pi/2,\pi/2)$ .

Watch out

If $xy>1$ the raw formula lands in the wrong branch, so you must correct it:

$\tan^{-1}x+\tan^{-1}y = \pi + \tan^{-1}\!\left(\frac{x+y}{1-xy}\right) \ (x,y>0),$

and $-\pi+\tan^{-1}(\cdot)$ when $x,y<0$ . Ignoring this is the classic exam slip.

Subtraction

Replacing $y$ with $-y$ gives

\tan^{-1}x - \tan^{-1}y = \tan^{-1}\!\left(\frac{x-y}{1+xy}\right), \qquad xy>-1.

The doubling identity

Set $y=x$ in the addition formula:

2\tan^{-1}x = \tan^{-1}\!\left(\frac{2x}{1-x^2}\right), \qquad -1<x<1.

The same angle can also be rewritten through sine and cosine, which is handy in integration and in proofs -- but each alternate form only matches $2\tan^{-1}x$ on part of its domain, so the two forms carry different conditions:

2\tan^{-1}x = \sin^{-1}\!\left(\frac{2x}{1+x^2}\right), \qquad -1\le x\le 1,

2\tan^{-1}x = \cos^{-1}\!\left(\frac{1-x^2}{1+x^2}\right), \qquad x\ge 0.

Watch out

The $\cos^{-1}$ form needs $x\ge 0$ -- it fails for negative $x$ . Check $x=-1$ : $2\tan^{-1}(-1)=2\left(-\tfrac{\pi}{4}\right)=-\tfrac{\pi}{2}$ , but $\cos^{-1}\!\left(\tfrac{1-1}{1+1}\right)=\cos^{-1}(0)=\tfrac{\pi}{2}$ , the wrong sign entirely. The $\sin^{-1}$ form has no such restriction because $\sin^{-1}$ (unlike $\cos^{-1}$ ) can return a negative angle.

The complementary identity

For every real $x$ ,

\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}.

This holds without restriction because $\tan^{-1}$ and $\cot^{-1}$ of the same value are complementary angles.

Tip

For $x>0$ : $\tan^{-1}\dfrac{1}{x} = \cot^{-1}x = \dfrac{\pi}{2}-\tan^{-1}x$ . For $x<0$ : $\tan^{-1}\dfrac{1}{x} = -\dfrac{\pi}{2}-\tan^{-1}x$ , but this is not the same as $\cot^{-1}x$ -- since $\cot^{-1}x$ always lies in $(0,\pi)$ (never negative), for $x<0$ it instead equals $\pi+\tan^{-1}\dfrac{1}{x}$ . Check $x=-1$ : $\cot^{-1}(-1)=\tfrac{3\pi}{4}$ , while $\tan^{-1}\dfrac{1}{-1}=-\tfrac{\pi}{4}$ -- these clearly are not equal, so never carry the $x>0$ shortcut over to negative $x$ .

The takeaway

Every inverse-tangent identity is the tangent addition formula read backwards. Learn the addition rule and its $xy$ conditions, then the subtraction, doubling, and complementary forms follow -- but always check the domain restriction on each alternate form before quoting it, since $\sin^{-1}$ , $\cos^{-1}$ and $\cot^{-1}$ each carry their own principal-range limits. That sign condition is where marks are won or lost.

The addition, subtraction, and doubling identities for tan⁻¹x are an important part of the CBSE Class 12 Inverse Trigonometric Functions chapter, and "tan inverse x plus tan inverse y formula with conditions" is a frequently searched topic because of the easy-to-miss xy conditions involved. These identities are tested regularly in both CBSE board exams and JEE Main inverse trigonometry problems.

We want the derivative of $\tan^{-1} x$ with respect to $\log x$ , not with respect to $x$ . Using the chain rule in parametric form, the answer is $\frac{x}{1+x^2}$ .

The key idea here is that "derivative with respect to $\log x$ " is not the same as the ordinary derivative $\frac{d}{dx}$ . When we say "differentiate $u$ with respect to $v$ ", we mean $\frac{du}{dv}$ . So we need $\frac{d(\tan^{-1} x)}{d(\log x)}$ .

This is a classic case of parametric differentiation: if both $u$ and $v$ are functions of $x$ , then

\frac{du}{dv} = \frac{du/dx}{dv/dx}.

Let’s apply that.

Let $u = \tan^{-1} x$ and $v = \log x$ . We want $\frac{du}{dv}$ .
First, find $\frac{du}{dx}$ :

\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}.

This is a standard result — the derivative of the inverse tangent.

Next, find $\frac{dv}{dx}$ :

\frac{d}{dx}(\log x) = \frac{1}{x}.

Since $x > 1$ , $\log x$ is the natural logarithm (base $e$ ), and its derivative is $1/x$ .

Now, using the parametric formula:

\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{\frac{1}{1+x^2}}{\frac{1}{x}} = \frac{1}{1+x^2} \cdot \frac{x}{1} = \frac{x}{1+x^2}.

Watch out

A common mistake is to compute $\frac{d}{dx}(\tan^{-1} x)$ and stop, forgetting that the question asks for the derivative with respect to $\log x$ , not $x$ . Always check the "denominator" of the derivative.

Tip

If you ever see "derivative of $f(x)$ with respect to $g(x)$ ", immediately think $\frac{f'(x)}{g'(x)}$ — provided $g'(x) \neq 0$ . This works because both are functions of the same underlying variable $x$ .

✓Final answer

\tan^{-1} x

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 39 on this concept.

CBSE 2026Set A1 markMCQ
$\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) =$
(a) $\frac{\pi}{3}$
(b) $\frac{\pi}{6}$
(c) $-\frac{\pi}{3}$
(d) $-\frac{\pi}{6}$
›Reveal solutionSolution

$\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}$ .

The principal value of $\tan^{-1}$ lies in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ .

We need the angle $\theta$ in this range with $\tan\theta = -\frac{1}{\sqrt{3}}$ .

Since $\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$ and $\tan$ is odd, $\tan\left(-\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}}$ .

And $-\frac{\pi}{6} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ , so it is the principal value.

✓Final answer
$-\frac{\pi}{6}$
CBSE 2026Set A1 markMCQ
$2\tan^{-1}\frac{1}{3} =$
(a) $\tan^{-1}\frac{3}{2}$
(b) $\tan^{-1}\frac{3}{4}$
(c) $\tan^{-1}\frac{4}{3}$
(d) $\tan^{-1}\frac{2}{3}$
›Reveal solutionSolution

$2\tan^{-1}\frac{1}{3} = \tan^{-1}\frac{3}{4}$ .

Use the double-angle identity valid for $|x|<1$ :

$2\tan^{-1}x = \tan^{-1}\frac{2x}{1-x^2}.$

Here $x = \frac{1}{3}$ :

$\frac{2\cdot\frac{1}{3}}{1-\frac{1}{9}} = \frac{\frac{2}{3}}{\frac{8}{9}} = \frac{2}{3}\cdot\frac{9}{8} = \frac{18}{24} = \frac{3}{4}.$

Hence $2\tan^{-1}\frac{1}{3} = \tan^{-1}\frac{3}{4}$ .

✓Final answer
$\tan^{-1}\frac{3}{4}$
CBSE 2026Set A1 markMCQ
$x \in R,\ \cot(\tan^{-1}x + \cot^{-1}x) =$
(a) $1$
(b) $\frac{1}{2}$
(c) $0$
(d) $\frac{1}{3}$
›Reveal solutionSolution

$\cot\left(\tan^{-1}x + \cot^{-1}x\right) = \cot\frac{\pi}{2} = 0$ .

For all real $x$ , the complementary identity gives

$\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}.$

Therefore

$\cot\left(\tan^{-1}x + \cot^{-1}x\right) = \cot\frac{\pi}{2} = 0.$

✓Final answer
$0$
CBSE 2026Set A1 markMCQ
$\tan^{-1}2 + \tan^{-1}3 =$
(a) $-\frac{\pi}{4}$
(b) $\frac{\pi}{4}$
(c) $\frac{3\pi}{4}$
(d) $\pi$
›Reveal solutionSolution

$\tan^{-1}2 + \tan^{-1}3 = \frac{3\pi}{4}$ .

Use the addition formula. With $a=2,\ b=3$ we have $ab=6>1$ , so

$\tan^{-1}a + \tan^{-1}b = \pi + \tan^{-1}\frac{a+b}{1-ab}.$

Compute:

$\frac{a+b}{1-ab} = \frac{5}{1-6} = \frac{5}{-5} = -1.$

So

$\tan^{-1}2 + \tan^{-1}3 = \pi + \tan^{-1}(-1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}.$

✓Final answer
$\frac{3\pi}{4}$
CBSE 2026Set A1 markMCQ
$\tan^{-1}\frac{x}{y} - \tan^{-1}\frac{x-y}{x+y} =$
(a) $-\frac{3\pi}{4}$
(b) $\frac{\pi}{2}$
(c) $\frac{\pi}{4}$
(d) $\frac{\pi}{3}$
›Reveal solutionSolution

$\tan^{-1}\frac{x}{y} - \tan^{-1}\frac{x-y}{x+y} = \frac{\pi}{4}$ .

Use $\tan^{-1}a - \tan^{-1}b = \tan^{-1}\frac{a-b}{1+ab}$ with $a=\frac{x}{y}$ , $b=\frac{x-y}{x+y}$ .

Numerator:

$a-b = \frac{x}{y} - \frac{x-y}{x+y} = \frac{x(x+y) - y(x-y)}{y(x+y)} = \frac{x^2+xy-xy+y^2}{y(x+y)} = \frac{x^2+y^2}{y(x+y)}.$

Denominator:

$1+ab = 1 + \frac{x}{y}\cdot\frac{x-y}{x+y} = \frac{y(x+y)+x(x-y)}{y(x+y)} = \frac{xy+y^2+x^2-xy}{y(x+y)} = \frac{x^2+y^2}{y(x+y)}.$

So the ratio is $1$ , giving

$\tan^{-1}(1) = \frac{\pi}{4}.$

✓Final answer
$\frac{\pi}{4}$
CBSE 2026Set A1 markMCQ
$|x| \le 1,\ \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) =$
(a) $2\cos^{-1}x$
(b) $2\sin^{-1}x$
(c) $2\tan^{-1}x$
(d) $\tan^{-1}2x$
›Reveal solutionSolution

$\cos^{-1}\frac{1-x^2}{1+x^2} = 2\tan^{-1}x$ (for $0\le x\le 1$ ).

Put $x = \tan\theta$ , so $\theta = \tan^{-1}x$ . Then

$\frac{1-x^2}{1+x^2} = \frac{1-\tan^2\theta}{1+\tan^2\theta} = \cos 2\theta.$

Therefore

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

✓Final answer
$2\tan^{-1}x$
CBSE 2025Set E1 markMCQ
$\cot^{-1}\left(\tan\frac{\pi}{7}\right) =$
(a) $\frac{\pi}{7}$
(b) $\frac{5\pi}{14}$
(c) $\frac{9\pi}{14}$
(d) $\frac{3\pi}{14}$
›Reveal solutionSolution

Convert the tangent to a cotangent using complementary angles; the answer is $\frac{5\pi}{14}$ .

Use $\tan\theta=\cot\left(\frac{\pi}{2}-\theta\right)$ :

$\tan\frac{\pi}{7}=\cot\left(\frac{\pi}{2}-\frac{\pi}{7}\right)=\cot\frac{7\pi-2\pi}{14}=\cot\frac{5\pi}{14}.$

Since $\frac{5\pi}{14}$ lies in the principal range $(0,\pi)$ of $\cot^{-1}$ ,

$\cot^{-1}\left(\tan\frac{\pi}{7}\right)=\cot^{-1}\left(\cot\frac{5\pi}{14}\right)=\frac{5\pi}{14}.$

✓Final answer
$\frac{5\pi}{14}$
CBSE 2025Set E1 markMCQ
$\tan^{-1}(-\sqrt{3}) =$
(a) $\frac{\pi}{6}$
(b) $\frac{\pi}{3}$
(c) $\frac{2\pi}{3}$
(d) $-\frac{\pi}{3}$
›Reveal solutionSolution

$\tan^{-1}$ is an odd function with principal range $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ ; the value is $-\frac{\pi}{3}$ .

Since $\tan\frac{\pi}{3}=\sqrt{3}$ and $\tan^{-1}(-x)=-\tan^{-1}x$ ,

$\tan^{-1}(-\sqrt{3})=-\tan^{-1}(\sqrt{3})=-\frac{\pi}{3}.$

This lies in the principal range, so it is the required value.

✓Final answer
$-\frac{\pi}{3}$
CBSE 2025Set E1 markMCQ
$\tan^{-1}(\sqrt{3}) - \cot^{-1}(-\sqrt{3}) =$
(a) $0$
(b) $-\frac{\pi}{2}$
(c) $\pi$
(d) $\frac{\pi}{2}$
›Reveal solutionSolution

Evaluate each inverse function in its principal range and subtract; result $-\frac{\pi}{2}$ .

First, $\tan^{-1}(\sqrt{3})=\frac{\pi}{3}$ .

For $\cot^{-1}(-\sqrt{3})$ , the principal range of $\cot^{-1}$ is $(0,\pi)$ . We need $\cot\theta=-\sqrt{3}$ with $\theta\in(0,\pi)$ . Since $\cot\frac{\pi}{6}=\sqrt{3}$ ,

$\cot^{-1}(-\sqrt{3})=\pi-\frac{\pi}{6}=\frac{5\pi}{6}.$

Therefore

$\tan^{-1}(\sqrt{3})-\cot^{-1}(-\sqrt{3})=\frac{\pi}{3}-\frac{5\pi}{6}=\frac{2\pi-5\pi}{6}=-\frac{3\pi}{6}=-\frac{\pi}{2}.$

✓Final answer
$-\frac{\pi}{2}$
CBSE 2025Set E1 markMCQ
$\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3} =$
(a) $\pi$
(b) $\frac{\pi}{4}$
(c) $\frac{\pi}{2}$
(d) $\frac{\pi}{3}$
›Reveal solutionSolution

Use the sum formula for inverse tangents; the sum is $\frac{\pi}{4}$ .

When $xy<1$ , $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\dfrac{x+y}{1-xy}$ . Here $xy=\frac16<1$ , so

$\tan^{-1}\frac12+\tan^{-1}\frac13=\tan^{-1}\frac{\frac12+\frac13}{1-\frac12\cdot\frac13}=\tan^{-1}\frac{\frac{5}{6}}{\frac{5}{6}}=\tan^{-1}1=\frac{\pi}{4}.$

✓Final answer
$\frac{\pi}{4}$
CBSE 2025Set E1 markMCQ
$\tan\left\{\frac{1}{2}\left(\tan^{-1}x + \tan^{-1}\frac{1}{x}\right)\right\} =$
(a) $1$
(b) $\sqrt{3}$
(c) $0$
(d) $\infty$
›Reveal solutionSolution

$\tan^{-1}x + \tan^{-1}\frac1x = \frac{\pi}{2}$ ; half is $\frac{\pi}{4}$ ; $\tan\frac{\pi}{4} = 1$ .

For $x > 0$ there is a standard identity:

$\tan^{-1}x + \tan^{-1}\frac{1}{x} = \frac{\pi}{2}.$

Taking half:

$\frac{1}{2}\left(\tan^{-1}x + \tan^{-1}\frac{1}{x}\right) = \frac{\pi}{4}.$

Hence

$\tan\left(\frac{\pi}{4}\right) = 1.$

✓Final answer
$1$
CBSE 2025Set ANNUAL1 markMCQ
$x$
(i) $\dfrac{1}{\sqrt3}$
(ii) $\dfrac13$
(iii) $\dfrac14$
(iv) $\dfrac{1}{\sqrt5}$
›Reveal solutionSolution

Simplify the RHS first, then solve for $x$ .

$\sin^{-1}\left(\frac{\sqrt3}{2}\right) = \frac{\pi}{3}$

So the equation becomes:

$2\tan^{-1}x = \frac{\pi}{3} \implies \tan^{-1}x = \frac{\pi}{6}$

$x = \tan\frac{\pi}{6} = \frac{1}{\sqrt3}$

✓Final answer
$x=\dfrac{1}{\sqrt3}$

Q.Find the derivative of tan⁡−1x\tan^{-1} xtan−1x with respect to log⁡x\log xlogx; (where x∈(1,∞)x \in (1, \infty)x∈(1,∞)).

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Q.Find the derivative of $\tan^{-1} x$ with respect to $\log x$ ; (where $x \in (1, \infty)$ ).

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