Question

# Let $$f(x)$$ be a differentiable function which satisfies the equation $$\displaystyle f(xy)=f(x)+f(y)$$ for all $$x>0$$,$$y>0$$ then $$f'(x)$$ is equal to

A
f(x)x
B
1x
C
f(x)
D
f(x).(lnx)

Solution

## The correct option is A $$\displaystyle\dfrac{f'(x)}{x}$$$$f\left( xy \right) =f\left( x \right) \neq f\left( y \right)$$Partial differentiation  w.r.t. x$$\partial \cfrac { f\left( xy \right) }{ \partial x } =\cfrac { \partial f\left( x \right) }{ \partial x } +\cfrac { \partial f\left( y \right) }{ \partial x }$$$$=\left( f^{ ' }\left( xy \right) \right) \times \left( y\cdot 1 \right) =f^{ ' }\left( x \right) +0$$Put $$y=\cfrac { 1 }{ x }$$$$\Rightarrow f^{ ' }\left( x \right) =f^{ ' }\left( x\times \cfrac { 1 }{ x } \right) \times \left( \cfrac { 1 }{ x } \right)$$$$=\cfrac { f^{ ' }\left( x \right) }{ x }$$Mathematics

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