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

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(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

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$2x^2$

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(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 $f'(x) =$

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