MCQ
Let $f(x)$ be a differentiable function which satisfies the equation $f(xy) = f(x) + f(y)$ for all $x > 0, y > 0$ then $f ‘(x)$ is equal to
  • $\frac{{f'(1)}}{x}$
  • B
    $\frac{1}{x}$
  • C
    $f ‘(1)$
  • D
    $f ‘(1)\cdot (lnx)$

Answer

Correct option: A.
$\frac{{f'(1)}}{x}$
a
$f ‘(x) =$ $\mathop {Limit}\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$

 $=\mathop {Limit}\limits_{h \to 0} \frac{{f\left( {x.\left( {1 + \frac{h}{x}} \right)} \right) - f(x)}}{h}$
$=\mathop {Limit}\limits_{h \to 0} \frac{{f\left( x \right) + f\left( {1 + \frac{h}{x}} \right) - f(x)}}{h}$ 

$=\mathop {Limit}\limits_{h \to 0} \frac{{f\left( {1 + \frac{h}{x}} \right)}}{{x\frac{h}{x}}}$ 
 $=\frac{1}{x}\mathop {Limit}\limits_{t \to 0} \frac{{f\left( {1 + t} \right) - f(1)}}{t}$ (note that $f(1) = 0$)

= $\frac{{f'(1)}}{x}$ 

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