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

1. zepp

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

2. saifoo.khan

Zippy! Me meet again. ;D

3. zepp

ohai saif :D

4. saifoo.khan

not sure how to solve this. :/

5. AccessDenied

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...

6. AccessDenied

\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*}

7. tkhunny

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.