derivative of an integral

$$\frac{d}{dx}\int f(x^2)dx^2$$

Question

derivative of an integral

$$\frac{d}{dx}\int f(x^2)dx^2$$

zepp · Answer

i've never seen a $\large dx^2$, anyone can explain to me what it is?

saifoo.khan · Answer

Zippy! Me meet again. ;D

zepp · Answer

ohai saif :D

saifoo.khan · Answer

not sure how to solve this. :/

accessdenied · Answer

I recommend attempting the substitution: $\sqrt{u} = x$.  Just a hunch that it may turn out well (and not because I just explained the solution on irc chats).  I think it may be helpful to our problem so that the f(x^2) dx^2 becomes f(u) du, etc...

accessdenied · Answer

$$
\newcommand\dd[1]{\,\mathrm d#1}
\newcommand\de[1]{\frac{\mathrm d}{\mathrm d#1}}

\text{Let } \sqrt{u} = x \text{.  Then, } \
\; \frac{1}{2 \sqrt{u}} \dd{u} = \dd{x} \
\begin{align*}
\implies \de{x} \int f(x^2) \dd{x^2}  &= \frac{\text{d}}{\left(\frac{1}{2\sqrt{u}}\right) \; \dd{u}} \int f(u) \; \dd{u} \
&= 2\sqrt{u} \de{u} \int f(u) \; \dd{u} \
&= 2x f(u) \
&= 2x f(x^2) \
\end{align*} $$

tkhunny · Answer

Another Way:

$\de{x}\int f(x^{2})\;dx^{2} = \de{x}\int  f(x^{2})\cdot 2x\;dx = \de{x} \left(F(x^{2}) + C\right) = f(x^{2})\cdot 2x$

Same thing, really.