Solution needed

Calculus Level 4

Let $f$ be a differentiable function satisfying the condition $f\left(\frac{x}{y}\right) = \dfrac{f(x)}{f(y)}$ for all $x,y$ .

If $f'(1)=2$ ,

Find $f'(x)$ .

Details and Assumptions :-

$f'(x)=\dfrac{d(f(x))}{dx}$ and $f'(k)$ is the value of $\frac{d(f(x))}{dx}$ at $x=k$ .

\dfrac{f(x)}{x}

2xf(x)

2f(x)

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

1 solution

Aditya Raut
Jul 24, 2015

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

Put $x=y$ to obtain $f(1)=1$ .

Now take partial derivative w.r.t. $x$ , i.e. consider $y$ as a constant.

$\displaystyle \therefore \frac{1}{y} \cdot f'\left(\frac{x}{y}\right) = \dfrac{f'(x)}{f(y)}$

Take partial derivative w.r.t. $y$ , i.e. consider $x$ as a constant.

$\displaystyle \therefore \frac{-x}{y^2} f'\left(\frac{x}{y}\right) = \dfrac{-f(x)f'(y)}{f(y)f(y)}$

Dividing the second equation by first equation,

$\displaystyle \dfrac{\frac{-x}{y^2}f'\left(\frac{x}{y}\right)}{\frac{1}{y}f'\left( \frac{x}{y}\right)} = \dfrac{\frac{-f(x)f'(y)}{f(y)f(y)}}{\frac{f'(x)}{f(y)}}$

$\therefore \dfrac{x}{y} = \dfrac{f(x)f'(y)}{f(y)f'(x)}$

$\therefore \dfrac{x f'(x)}{f(x)}=\dfrac{yf'(y)}{f(y)}$

Now just put $y=1$ , we get

$\therefore \dfrac{xf'(x)}{f(x)}=\dfrac{1\times 2}{1} =2$

$\therefore \boxed{f'(x)=\dfrac{2f(x)}{x}}$

@jaikirat sandhu , a solution, as you had asked. :)

Aditya Raut - 5 years, 10 months ago

Solution needed

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