Derivative of x^3 + 1 with respect to x^2 + 1 is Option 1: 2x/3 Option 2: x/3 Option 3: x/2 Option 4: 3x/2

Answer: 3x/2. Solution: Let u = x^3+1 and v = x^2+1. du/dv = (du/dx)/(dv/dx) = 3x^2/2x = 3x/2.

Calculus

Continuity & Differentiability

Easy

Derivative of $x^3 + 1$ with respect to $x^2 + 1$ is

$\frac{2x}{3}$

$\frac{x}{3}$

$\frac{x}{2}$

$\frac{3x}{2}$

✅ Correct Option: 4

Let $u = x^3+1$ and $v = x^2+1$ .

$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{3x^2}{2x} = \frac{3x}{2}$ .

2025: 16 May Shift 1

Match List-I with List-II

List-I	List-II
(A) $f(x) = x \sin x$	(I) is not continuous at $x = -3$
(B) $f(x) = \frac{\vert x\vert }{x}, x \neq 0$ and $f(x) = 1 \text{ at } x = 0$	(II) is continuous everywhere
(C) $f(x) = x - [x]$ , $[x]$ denotes greatest integer function	(III) is not differentiable at $x = 1$
(D) $f(x) = e^{\vert x - 1\vert }$	(IV) is not continuous at $x = 0$

Choose the correct answer from the options given below:

2025: 19 May Shift 1

If $x = e^{\cos 2t}$ , $y = e^{\sin 2t}$ , then $\frac{dy}{dx}$ equals to

2023: 30 May Shift 3

The derivative of $\sec(\tan \sqrt{x})$ with respect to x is :

2026: 29 May Shift 2

The function $f(x)=\begin{cases}\frac{e^{3x}-e^{-5x}}{x}, & x\neq 0\\ \lambda, & x=0\end{cases}$ is continuous at x=0 for the value of ' $\lambda$ ' as:

2026: 25 May Shift 1

Match the LIST-I with LIST-II

LIST-I	LIST-II
A. $f(x) = \log_e(x^2+1)$	I. not continuous at all integral values
B. $f(x) = \lvert x\rvert$	II. is continuous as well as differentiable at $x=0$
C. $f(x) = x^2$	III. Continuous everywhere
D. $f(x) = [x]$ , $[x]$ is greatest integer function	IV. Continuous everywhere but not differentiable at $x=0$

Choose the correct answer from the options given below:

2026: 29 May Shift 2

If $x=a(1-\cos\theta), y=a(\theta+\sin\theta)$ then $\frac{d^2y}{dx^2}$ is equal to

Derivative of x3+1x^3 + 1x3+1 with respect to x2+1x^2 + 1x2+1 is

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

Derivative of $x^3 + 1$ with respect to $x^2 + 1$ is