Let f: R rightarrow R be a differentiable fun | vedclass

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(C) Given the functional equation $f(x+y)=f(x) \cdot f^{\prime}(y)+f^{\prime}(x) \cdot f(y)$.Putting $x=0$ and $y=0$,we get $f(0)=f(0) \cdot f^{\prime

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$2$

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(C) Given the functional equation $f(x+y)=f(x) \cdot f^{\prime}(y)+f^{\prime}(x) \cdot f(y)$.Putting $x=0$ and $y=0$,we get $f(0)=f(0) \cdot f^{\prime}(0)+f^{\prime}(0) \cdot f(0) = 2 f(0) \cdot f^{\prime}(0)$.Since $f(0)=1$,we have $1 = 2(1) \cdot f^{\prime}(0)$,which implies $f^{\prime}(0)=\frac{1}{2}$.Now,keeping $x$ as a variable and setting $y=0$,we get $f(x+0)=f(x) \cdot f^{\prime}(0)+f^{\prime}(x) \cdot f(0)$.Substituting $f(0)=1$ and $f^{\prime}(0)=\frac{1}{2}$,we obtain $f(x) = f(x) \cdot \frac{1}{2} + f^{\prime}(x) \cdot 1$.This simplifies to $f^{\prime}(x) = \frac{1}{2} f(x)$,or $\frac{f^{\prime}(x)}{f(x)} = \frac{1}{2}$.Integrating both sides with respect to $x$,we get $\int \frac{f^{\prime}(x)}{f(x)} dx = \int \frac{1}{2} dx$,which results in $\log(f(x)) = \frac{1}{2}x + C$.Using $f(0)=1$,we have $\log(1) = 0 + C$,so $C=0$.Thus,$\log(f(x)) = \frac{1}{2}x$.For $x=4$,$\log(f(4)) = \frac{1}{2} 	imes 4 = 2$.

List-$I$	List-$II$
$A. x^2 + y^2 + 3xy = 7$	$I. \frac{x^2 + ay}{ax + y^2}$
$B. x^{2/3} + y^{2/3} = a^{2/3}$	$II. \frac{-(2x + 3y)}{3x + 2y}$
$C. x^3 + y^3 = 3axy$	$III. -(\frac{y}{x})^{1/3}$
$D. xy(x - y) = 2$	$IV. \frac{x^2 - ay}{ax - y^2}$
	$V. \frac{-y(2x + y)}{x(x + 2y)}$

Let $f: R \rightarrow R$ be a differentiable function with $f(0)=1$ and satisfying the equation $f(x+y)=f(x) \cdot f^{\prime}(y)+f^{\prime}(x) \cdot f(y)$ for all $x, y \in R$. Then the value of $\log (f(4))$ is

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