A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
Calculus! (See comments)
anonymous
 3 years ago
Calculus! (See comments)

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac {d}{dx} \int\limits_{\cos 3x}^{\tan 3x} \sqrt{1+t^2}\] Preferably with steps+rationale shown. Thanks!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If the equation is too small to see: Find the derivative of the integral of squrt(1+t^2)dt from cos3x to tan 3x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0(Second fundamental therom)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Suppose you have the following: \[ F(x) = \int_a^{g(x)}f(t)dt \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So @mathg8 , so far in my class we have only solved problems with one of the limits of integration being an expression, and one would be a constant (ie 3), how would this differ.

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.1You would have to split the integral at some constant (say a)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Then suppose : \[ H(y) = \int_a^yf(t)dt \]Notice that \(H(y) = H(g(x)) = F(x)\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0By the fundamental theorem: \[ \frac{d}{dy}H(y) = f(y) = f(g(x)) \]

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.1\[\frac {d}{dx} \int\limits\limits_{\cos 3x}^{\tan 3x} \sqrt{1+t^2}dt = \frac {d}{dx} \int\limits\limits_{\cos 3x}^{a} \sqrt{1+t^2}dt+\frac {d}{dx} \int\limits\limits_{a}^{\tan 3x} \sqrt{1+t^2}dt\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0See where I'm going with this?

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.1and \[\int\limits\limits_{\cos 3x}^{a} \sqrt{1+t^2}=\int\limits\limits_{a}^{\cos3x} \sqrt{1+t^2}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@wio I'm reading through it (:

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.1If you know how to work with an expression other than just "x" on the upper integration bound, then you should have no problem computing it the way I presented it by splitting your integral into 2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Basically we have: \[ \frac{d}{dy}F(x) = f(g(x)) \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@kirbykirby Ok, thanks!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@wio here, let me put what you said in words to see if I get it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Argonx16 I'm not finished yet but sure.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If you have f(x) = integral of f(t) from a to g(x), and y=g(x), and \[H(y)=\int\limits_{a}^{y}f(t)dt\] Then we can see that H(y)=H(g(x)) because of the fact that y=g(x) (Not quite sure how H(y)=H(g(x)) though) @wio

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0When we say \(y=g(x)\) we're just saying they are interchangeable.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The reason I did that was to make it so you can see that a function is sort of just like a variable.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0also there is a difference between \(F(x)\) and \(f(x)\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry, my reply was written wrong. Let me restate it.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I can see that H(y) = H (g(x)) (y=g(x)), but I /don't/ see how H(y)=F(x) (or H(g(x)=F(x))

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It's the transitive property of equality. \[ H(y)=\int\limits_{a}^{y}f(t)dt = \int\limits_{a}^{g(x)}f(t)dt =F(x) \]I'm not saying they are the same function though. I'm just saying they are equal.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Look if you want, just ignore \(H(y)\). If I said: \[ F(x)=\int\limits_{a}^{y}f(t)dt \]Do you understand that by the fundamental theorem: \[ \frac{d}{dy}F(x) = f(y) \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ok, I see what you did to make F(x)=H(y)=H(g(x)) And, yes I understand the second fundamental theorem.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now the trick here is to multiply both sides by \(dy/dx\) \[ \frac{d}{dy}F(x) \frac{dy}{dx} = f(y)\frac{dy}{dx} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0By the chain rule: \[ \frac{d}{dy}F(x) \frac{dy}{dx} = \frac{d}{dx}F(x) = f(y)\frac{dy}{dx} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Argonx16 this is sort of the tricky part, but do you get it?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Another way to write this is: \[ \frac{dF(x)}{dy}\frac{dy}{dx} = \frac{dF(x)}{dx} \]Or \[ \frac{dF}{dy}\frac{dy}{dx} = \frac{dF}{dx} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes I got the chain rule step (esp after you rewrote it in a different way).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So we got:\[ \frac{d}{dx}F(x) = \frac{d}{dx}\int\limits_{a}^{y}f(t)dt = \frac{d}{dx}\int\limits_{a}^{g(x)}f(t)dt \]And: \[ f(y)\frac{dy}{dx} = f(g(x))\frac{dg(x)}{dx} = f(g(x))g'(x) \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Putting it all together: \[ \frac{d}{dx} \left[ \int\limits_{a}^{g(x)}f(t)dt \right] = f(g(x))g′(x) \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ \frac {d}{dx} \int\limits_{\cos 3x}^{\tan 3x} \sqrt{1+t^2} = \frac {d}{dx} \int\limits_{z}^{\tan 3x} \sqrt{1+t^2} \frac {d}{dx} \int\limits^{\cos 3x}_{a} \sqrt{1+t^2} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0To expand on the earlier realization: \[ \frac{d}{dx} \left[ \ \int\limits_{h(x)}^{g(x)}f(t)dt \right] = f(g(x))g′(x) f(h(x))h'(x) \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Argonx16 Does it all make sense now?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes! @wio Thanks for spending so much time helping me on this! I know how much patience/time/effort it takes to help so in depth, so I am very appreciative of your help :D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I've done this thing like 2 or 3 times and every time I it gets a bit easier to teach.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah. It definitely helps so much more. OS is also a good way to retain math concepts taught earlier :D
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.