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

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

Let

f

be a differentiable function satisfying the condition:

f\left( {\frac{x}{y}} \right) = \frac{{f(x)}}{{f(y)}}\mathrm{ }\,\,\mathrm{for}\,\,\mathrm{ all}\,\,\mathrm{ }x,y( \ne 0) \in R\mathrm{ }\,\,\mathrm{and }\,\,f(y) \ne 0

f^{\prime}(1)=2

, then

f^{\prime}(x)

is equal to

View answer and explanation

Correct option: D

\frac{2 f(x)}{x}

Option D is correct because using the given condition

f\left( {\frac{x}{y}} \right) = \frac{{f(x)}}{{f(y)}}

and

f^{\prime}(1)=2

, we find

f^{\prime}(x) = \frac{2 f(x)}{x}