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 equatio

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

The derivative of ${x^6} + {6^x}$ with respect to $x$ is:

If $f(x) = \sqrt{x + 2 \sqrt{2x - 4}} + \sqrt{x - 2 \sqrt{2x - 4}}$,then the value of $10 \times f^{\prime}(102) =$

For the differentiable function $f: R - \{0\} \rightarrow R$,let $3 f(x) + 2 f\left(\frac{1}{x}\right) = \frac{1}{x} - 10$. Then $\left|f(3) + f^{\prime}\left(\frac{1}{4}\right)\right|$ is equal to

If $y = \sin^{-1} \left( x\sqrt{1 - x} + \sqrt{x} \sqrt{1 - x^2} \right)$ and $\frac{dy}{dx} = \frac{1}{2\sqrt{x(1 - x)}} + p$,then $p =$

If $f(x) = (\log _{\cot x}\tan x)(\log _{\tan x}\cot x)^{-1}$,then $f'(2) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module