A community for students.
Here's the question you clicked on:
 0 viewing
zepp
 2 years ago
derivative of an integral
\[\frac{d}{dx}\int f(x^2)dx^2\]
zepp
 2 years ago
derivative of an integral \[\frac{d}{dx}\int f(x^2)dx^2\]

This Question is Closed

zepp
 2 years ago
Best ResponseYou've already chosen the best response.0i've never seen a \(\large dx^2\), anyone can explain to me what it is?

saifoo.khan
 2 years ago
Best ResponseYou've already chosen the best response.0Zippy! Me meet again. ;D

saifoo.khan
 2 years ago
Best ResponseYou've already chosen the best response.0not sure how to solve this. :/

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.1I 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
 2 years ago
Best ResponseYou've already chosen the best response.1\[ \newcommand\dd[1]{\,\mathrm d#1} \newcommand\de[1]{\frac{\mathrm d}{\mathrm d#1}} \text{Let } \sqrt{u} = x \text{. Then, } \\ \; \frac{1}{2 \sqrt{u}} \dd{u} = \dd{x} \\ \begin{align*} \implies \de{x} \int f(x^2) \dd{x^2} &= \frac{\text{d}}{\left(\frac{1}{2\sqrt{u}}\right) \; \dd{u}} \int f(u) \; \dd{u} \\ &= 2\sqrt{u} \de{u} \int f(u) \; \dd{u} \\ &= 2x f(u) \\ &= 2x f(x^2) \\ \end{align*} \]

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1Another 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.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.