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