If xy_e(x+y) = 1, then dy/dx = Option 1: -(1/x^2+y/(x+y))(y/(x+y)+_e(x+y)) Option 2: (-(x+y+x^2y))/(x^2[y+(x+y)_e(x+y)]) Option 3: (1/x^2+y/(x+y))(y/(x+y)+_e(x+y)) Option 4: (x+y+x^2y)/(x^2[y+(x+y)_e(x+y)])

Answer: (-(x+y+x^2y))/(x^2[y+(x+y)_e(x+y)]). Solution: Differentiating implicitly, y_e(x+y)+xy/(x+y) + [x_e(x+y)+xy/(x+y)]dy/dx = 0. Using _e(x+y) = 1/xy from the given relation and multiplying through by (x+y), the numerator becomes (x+y+x^2y)/x and the denominator x[y+(x+y)_e(x+y)]. Hence dy/dx = (-(x+y+x^2y))/(x^2[y+(x+y)_e(x+y)]).

Calculus

Continuity & Differentiability

Hard

common

If $xy\log_e(x+y) = 1$ , then $\dfrac{dy}{dx} =$

$-\left(\dfrac{1}{x^2}+\dfrac{y}{x+y}\right)\left(\dfrac{y}{x+y}+\log_e(x+y)\right)$

$\dfrac{-(x+y+x^2y)}{x^2[y+(x+y)\log_e(x+y)]}$

$\left(\dfrac{1}{x^2}+\dfrac{y}{x+y}\right)\left(\dfrac{y}{x+y}+\log_e(x+y)\right)$

$\dfrac{(x+y+x^2y)}{x^2[y+(x+y)\log_e(x+y)]}$

✅ Correct Option: 2

Differentiating implicitly, $y\log_e(x+y)+\dfrac{xy}{x+y} + \left[x\log_e(x+y)+\dfrac{xy}{x+y}\right]\dfrac{dy}{dx} = 0$ .

Using $\log_e(x+y) = \dfrac{1}{xy}$ from the given relation and multiplying through by $(x+y)$ , the numerator becomes $\dfrac{x+y+x^2y}{x}$ and the denominator $x[y+(x+y)\log_e(x+y)]$ .

Hence $\dfrac{dy}{dx} = \dfrac{-(x+y+x^2y)}{x^2[y+(x+y)\log_e(x+y)]}$ .

2025: 13 May Shift 1

Match List-I with List-II

List-I	List-II
Function	Property
(A) $f(x) = \begin{cases} \frac{x}{\vert x \vert} & : x \neq 0 \\ 0 & : x = 0 \end{cases}$	(I) continuous but not differentiable at $x= 0$
(B) $f(x) = \vert x \vert$	(II) continuous but not differentiable at $x=1$
(C) $f(x) = \vert x^2 - 1 \vert$	(III) discontinuous at $x = 0$
(D) $f(x) = \vert x - 1 \vert$	(IV) continuous but not differentiable at $x = 1, -1$

Choose the correct answer from the options given below:

2026: 22 May Shift 1

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2025: 19 May Shift 1

The value of k for which the function $f(x) = \begin{cases} \frac{1-\cos 8x}{16x^2}, & \text{if } x \neq 0 \\ k, & \text{if } x = 0 \end{cases}$ is continuous at $x = 0$ is:

2023: 15 June Shift 2

The points of discontinuity of the function f defined by $f(x) = \begin{cases} x+2 & x \leq 1 \\ x-2 & 1 < x < 2 \\ 0 & x \geq 2 \end{cases}$ are :

2022: 25 May Shift 1

The derivative of $\sin(\tan^{-1} e^{2x})$ with respect to $x$ is:

2026: 15 May Shift 1

If $y = \log_e \sin(e^x + 5x + 500)$ then $\dfrac{dy}{dx}$ is equal to

If xylog⁡e(x+y)=1xy\log_e(x+y) = 1xyloge​(x+y)=1, then dydx=\dfrac{dy}{dx} =dxdy​=

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

If $xy\log_e(x+y) = 1$ , then $\dfrac{dy}{dx} =$