Consider the non-constant differentiable func | vedclass

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(A) Given the functional equation $\frac{f(x)}{f(y)}=f(x-y)$.Setting $y=0$, we get $\frac{f(x)}{f(0)}=f(x)$, which implies $f(0)=1$.Differentiating bo

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$\frac{p^{2}}{q}$

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(A) Given the functional equation $\frac{f(x)}{f(y)}=f(x-y)$.Setting $y=0$, we get $\frac{f(x)}{f(0)}=f(x)$, which implies $f(0)=1$.Differentiating both sides with respect to $x$, we get $\frac{f^{\prime}(x)}{f(y)}=f^{\prime}(x-y)$.Setting $x=0$, we have $\frac{f^{\prime}(0)}{f(y)}=f^{\prime}(-y)$.Since $f^{\prime}(0)=p$, we get $f^{\prime}(-y) = \frac{p}{f(y)}$.Also, differentiating the original equation with respect to $y$, we get $f(x) \cdot (-\frac{f^{\prime}(y)}{(f(y))^2}) = f^{\prime}(x-y) \cdot (-1)$.This simplifies to $\frac{f(x) f^{\prime}(y)}{(f(y))^2} = f^{\prime}(x-y)$.At $y=0$, $\frac{f(x) f^{\prime}(0)}{(f(0))^2} = f^{\prime}(x)$, so $f^{\prime}(x) = p f(x)$.This is a linear differential equation with solution $f(x) = e^{px}$.Then $f^{\prime}(x) = p e^{px}$.Given $f^{\prime}(5) = q$, we have $p e^{5p} = q$, so $e^{5p} = \frac{q}{p}$.We need $f^{\prime}(-5) = p e^{-5p} = \frac{p}{e^{5p}} = \frac{p}{q/p} = \frac{p^2}{q}$.

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

Consider the non-constant differentiable function $f$ of one variable which obeys the relation $\frac{f(x)}{f(y)}=f(x-y)$. If $f^{\prime}(0)=p$ and $f^{\prime}(5)=q$, then $f^{\prime}(-5)$ is

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