Given that $\frac{\mathrm{d} }{\mathrm{d} x}f(\frac{x}{2}) = 2x$, find the value of $\frac{\mathrm{d} }{\mathrm{d} x}f(x)$

Question

Given that $\frac{\mathrm{d} }{\mathrm{d} x}f(\frac{x}{2}) = 2x$, find the value of $\frac{\mathrm{d} }{\mathrm{d} x}f(x)$.

My attempt:

let $t = \frac{x}{2}$, so $\frac{\mathrm{d} }{\mathrm{d} x}f(t) = 2t$

$\Rightarrow \frac{\mathrm{d} }{\mathrm{d} x}f(x) = 2x$

Is this correct? I am not sure of my answer.

$\begingroup$ Do you know of the chain rule? $\endgroup$
– user304329
Commented Apr 11, 2017 at 13:39 — user304329, Commented Apr 11, 2017 at 13:39
$\begingroup$ Yes I do know it. $\endgroup$
– Aspiring Mathlete
Commented Apr 11, 2017 at 13:51 — Aspiring Mathlete, Commented Apr 11, 2017 at 13:51

user160110 · Accepted Answer · 2017-04-11 13:47:38Z

6

Define $g(x)=f(\frac{x}{2})$ then $\frac{d}{dx}g(x)=2x$. Integrating both sides $$\int \frac{d}{dx}g(x)dx =\int 2x dx$$

Because the antiderivative $\frac{d}{dx}g(x)$ is $g(x)$ (fundemental theorem of calculas) we see that this equation becomes: $$g(x)=x^2+c$$

Now remember $f(\frac{x}{2})=g(x)$ so $$f(x)=g(2x)=(2x)^2+c=4x^2+c$$

thus $\frac{d}{dx}f(x)=8x$

answered Apr 11, 2017 at 13:47

user160110

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$\begingroup$ doesn't $f(x) = 2x^2$ solve it? then f'(x) = 4x : f'(x/2) = 2x $\endgroup$
– Cato
Commented Apr 11, 2017 at 15:12
1

$\begingroup$ @Cato In that case $f(\frac{x}{2})= \frac{1}{2}(x)^2$ thus $f'(\frac{x}{2})=x$ $\endgroup$
– user160110
Commented Apr 11, 2017 at 15:21
$\begingroup$ if f(x) = 4x^2, then f'(1) = 8 and f'(1/2) = 4 : so for x = 1, you are getting f'(x) = 8x and f'(x/2) = 4x not 2x $\endgroup$
– Cato
Commented Apr 11, 2017 at 15:49
1

$\begingroup$ @Cato Changing the variable changes how derivatives work. For example $f(x)=\sin x$ and $g(x)=\sin 2x$. Clearly $g(x)=f(2x)$, but $g'(x)\neq f'(2x)$. $\endgroup$
– user160110
Commented Apr 11, 2017 at 16:26

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