A function f: R rightarrow R is such that y f | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the equation: $y f(x+y) + \cos(mxy) = 1 + y f(x)$.Rearranging the terms,we get: $y(f(x+y) - f(x)) = 1 - \cos(mxy)$.Given $m=2$,the equation

Vedclass · Accepted Answer

$2x^2$

Vedclass · Answer

(D) Given the equation: $y f(x+y) + \cos(mxy) = 1 + y f(x)$.Rearranging the terms,we get: $y(f(x+y) - f(x)) = 1 - \cos(mxy)$.Given $m=2$,the equation becomes: $y(f(x+y) - f(x)) = 1 - \cos(2xy)$.Dividing by $y^2$ on both sides (assuming $y 
eq 0$): $\frac{f(x+y) - f(x)}{y} = \frac{1 - \cos(2xy)}{y^2}$.Using the identity $1 - \cos(2	heta) = 2 \sin^2(	heta)$,we have: $\frac{f(x+y) - f(x)}{y} = \frac{2 \sin^2(xy)}{y^2}$.This can be written as: $\frac{f(x+y) - f(x)}{y} = 2 \left( \frac{\sin(xy)}{y} ight)^2$.To find $f'(x)$,take the limit as $y ightarrow 0$:$f'(x) = \lim_{y ightarrow 0} \frac{f(x+y) - f(x)}{y} = \lim_{y ightarrow 0} 2 \left( \frac{\sin(xy)}{y} ight)^2$.Multiply and divide by $x^2$: $f'(x) = 2x^2 \lim_{y ightarrow 0} \left( \frac{\sin(xy)}{xy} ight)^2$.Since $\lim_{	heta ightarrow 0} \frac{\sin 	heta}{	heta} = 1$,we get: $f'(x) = 2x^2(1)^2 = 2x^2$.

$A$ function $f: R \rightarrow R$ is such that $y f(x+y) + \cos(mxy) = 1 + y f(x)$. If $m=2$,then $f'(x) =$

Similar Questions

Let $f(x + y) = f(x)f(y)$ and $f(x) = 1 + \sin(3x)g(x)$,where $g(x)$ is continuous. Then $f'(x)$ is:

$\frac{d}{dx}(\sin^{-1}x)$ is equal to

The derivative of $\tan x - x$ with respect to $x$ is

If $y = \sqrt{\sin \sqrt{x}}$,then $\frac{dy}{dx} = $

If $y = \frac{x}{2}\sqrt{a^2 + x^2} + \frac{a^2}{2}\log(x + \sqrt{x^2 + a^2})$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module