A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 2 years ago
Find \(\frac{df}{dx}\) where \[f(x) = \int_{x}^{x^2}\frac{1}{\sqrt{1+t^6}}dt\] for x>0 How to start??
 2 years ago
Find \(\frac{df}{dx}\) where \[f(x) = \int_{x}^{x^2}\frac{1}{\sqrt{1+t^6}}dt\] for x>0 How to start??

This Question is Closed

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.2So this is one of those problems that involves requires you to apply the FTC Part 1. The Fundamental Theorem of Calculus part 1 is,\[\huge \frac{d}{dx}\int\limits_c^x f(t) dt=f(x)\] With this problem, we have an function of x within the upper and lower limits, so it'll look a bit different.\[\large \frac{d}{dx}\int\limits_{x}^{x^2}f(t)dt=f(x^2)\color{blue}{(\frac{d}{dx}x^2)}f(x)\color{blue}{(\frac{d}{dx}x)}\]The blue pieces are there because we have to apply the chain rule.

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1Fundamental Theorem of Integral Calculus will get you started. \(f(x) = \int\limits_{x}^{x^{2}}g(t)\;dt\;=\;G(x^{2})  G(x)\) Where \(G(x)\) is an antiderivative of \(g(x)\). Chain Rule will get us the rest of the way. \(\dfrac{d}{dx}\left(G(x^{2})  G(x)\right)\;=\;g(x^{2})\cdot (2x)  g(x)\cdot (1)\)

Callisto
 2 years ago
Best ResponseYou've already chosen the best response.0I'm sorry... I was so blind... Thanks all!!!!

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1Don't forget that we DO have to believe that the antiderivative EXISTS! Good work.

Callisto
 2 years ago
Best ResponseYou've already chosen the best response.0f(x) = f(x) too, I guess. Since it's an even function..
Ask your own question
Ask a QuestionFind 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.