Find the derivative

[Guest]

Find f'(x) if it is known that
\(\displaystyle \frac{d}{dx}[f(2x)] = x^2\)

I let u(x) = 2x, then
\(\displaystyle \frac{d}{dx}[f(u)] = \frac{dy}{du} \frac{du}{dx}\)
\(\displaystyle \frac{d}{dx}[f(2x)] = 2 \frac{dy}{du}\)
therefore
\(\displaystyle \frac{dy}{du} = \frac{1}{2}x^2\)
then
\(\displaystyle f'(x) = \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}\)
\(\displaystyle = (\frac{1}{2}x^2) (2)\)
\(\displaystyle = x^2\)
why doesn't this work??

 

[pka]

\(\displaystyle \L
\begin{array}{l}
\frac{d}{{dx}}\left( {f(2x)} \right) = x^2 \\
\frac{d}{{dx}}\left( {f(2x)} \right) = 2f'(2x)\quad \Rightarrow \quad f'(2x) = \frac{{x^2 }}{2} \\
u = 2x\quad \Rightarrow \quad f'(u) = \frac{{u^2 }}{8} \\
f'(x) = \frac{{x^2 }}{8} \\
\end{array}\)

 

[Guest]

I still don't understand why my method does not work.

 

[pka]

Didn't you just go in one large circle?

 

[Guest]

I still don't understand