Let $f$ be differentiable function satisfying the condition $f\left(\frac{x}{y}\right)=\frac{f\left(x\right)}{f\left(y\right)}$ for all $x,y(\ne 0)\in R$ and  $f\left(y\right)\ne 0.$ If  $f^1\left(1\right)=2$ then $f^1\left(x\right)$ is equal to

$$\begin{gathered} f\left(\frac{x}{y}\right)=\frac{f\left(x\right)}{f\left(y\right)}Putx=y=1 \\ {f}^1\left(x\right)=\underset{h\to 0}{lt}\frac{f(x+h)-f\left(x\right)}{h} \\ f\left(x\right).\underset{h\to 0}{lt}\frac{\frac{f(x+h)}{f\left(x\right)}-1}{h} \\ =f\left(x\right).\underset{h\to 0}{lt}\frac{f\left(\frac{x+h}{x}\right)-1}{h} \\ =\frac{f\left(x\right)}{x}.\underset{h\to 0}{lt}\frac{f(1+\frac{h}{x})-f\left(1\right)}{\frac{h}{x}} \\ =\frac{f\left(x\right)}{x}.f^1\left(0\right)=\frac{2f\left(x\right)}{x}\therefore f^1\left(1\right)=2 \end{gathered}$$