If y = x(a+y), then dy/dx = Option 1: (^2(a+y))/( a) Option 2: (^2(a+y))/( a) Option 3: (^2(a+y))/( a) Option 4: (^2(a+y))/( a)

Answer: (^2(a+y))/( a). Solution: From x = ( y)/((a+y)), differentiate w.r.t. y: dx/dy = (- y (a+y) + y (a+y))/(^2(a+y)) = ( a)/(^2(a+y)). So dy/dx = (^2(a+y))/( a).

Calculus

Continuity & Differentiability

Medium

core

If $\cos y = x\cos(a+y)$ , then $\frac{dy}{dx} =$

$\frac{\cos^2(a+y)}{\sin a}$

$\frac{\sin^2(a+y)}{\sin a}$

$\frac{\cos^2(a+y)}{\cos a}$

$\frac{\sin^2(a+y)}{\cos a}$

✅ Correct Option: 1

From $x = \frac{\cos y}{\cos(a+y)}$ , differentiate w.r.t. $y$ :

$\frac{dx}{dy} = \frac{-\sin y \cos(a+y) + \cos y \sin(a+y)}{\cos^2(a+y)} = \frac{\sin a}{\cos^2(a+y)}$ .

So $\frac{dy}{dx} = \frac{\cos^2(a+y)}{\sin a}$ .

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The second order derivative of which of the following functions is $5^{\mathrm{x}}$ ?

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If $x=4t^3$ and $y=6t^2$ then $\frac{d^2y}{dx^2}$ at $t=1$ is

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Let $x = t^2, y = t^3$ . Then $\frac{d^2y}{dx^2}$ is equal to

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Match List-I with List-II

List-I	List-II
Function	Derivative
---	---
(A) $y = \sin^{-1} x + \sin^{-1} \sqrt{1 - x^2}; \vert x\vert < 1$	(I) $\frac{dy}{dx} = \frac{1}{2y-1}$
(B) $y = \sqrt{x + y}, x+y > 0 \text{ and } y \neq \frac{1}{2}$	(II) $\frac{dy}{dx} = 10^x \log_e 10$
(C) $y = \log_{10} x, x > 0$	(III) $\frac{dy}{dx} = 0$
(D) $y = 10^x$	(IV) $\frac{dy}{dx} = \frac{1}{x \log_e 10}$

Choose the correct answer from the options given below:

If cos⁡y=xcos⁡(a+y)\cos y = x\cos(a+y)cosy=xcos(a+y), then dydx=\frac{dy}{dx} = dxdy​=

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CUET Mathematics Past Year Questions (Topic-Wise)

If $\cos y = x\cos(a+y)$ , then $\frac{dy}{dx} =$