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Argonx16 Group TitleBest 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!
 one year ago

Argonx16 Group TitleBest ResponseYou've already chosen the best response.0
If the equation is too small to see: Find the derivative of the integral of squrt(1+t^2)dt from cos3x to tan 3x
 one year ago

Argonx16 Group TitleBest ResponseYou've already chosen the best response.0
(Second fundamental therom)
 one year ago

mathg8 Group TitleBest ResponseYou've already chosen the best response.0
check FTC 2
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.3
Suppose you have the following: \[ F(x) = \int_a^{g(x)}f(t)dt \]
 one year ago

Argonx16 Group TitleBest ResponseYou've already chosen the best response.0
So @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.
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.3
Let \(y = g(x)\)
 one year ago

kirbykirby Group TitleBest ResponseYou've already chosen the best response.1
You would have to split the integral at some constant (say a)
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.3
Then suppose : \[ H(y) = \int_a^yf(t)dt \]Notice that \(H(y) = H(g(x)) = F(x)\)
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.3
By the fundamental theorem: \[ \frac{d}{dy}H(y) = f(y) = f(g(x)) \]
 one year ago

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

wio Group TitleBest ResponseYou've already chosen the best response.3
See where I'm going with this?
 one year ago

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

Argonx16 Group TitleBest ResponseYou've already chosen the best response.0
@wio I'm reading through it (:
 one year ago

kirbykirby Group TitleBest ResponseYou've already chosen the best response.1
If 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
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.3
Basically we have: \[ \frac{d}{dy}F(x) = f(g(x)) \]
 one year ago

Argonx16 Group TitleBest ResponseYou've already chosen the best response.0
@kirbykirby Ok, thanks!
 one year ago

Argonx16 Group TitleBest ResponseYou've already chosen the best response.0
@wio here, let me put what you said in words to see if I get it
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.3
@Argonx16 I'm not finished yet but sure.
 one year ago

Argonx16 Group TitleBest ResponseYou've already chosen the best response.0
If 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
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.3
When we say \(y=g(x)\) we're just saying they are interchangeable.
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.3
The reason I did that was to make it so you can see that a function is sort of just like a variable.
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.3
also there is a difference between \(F(x)\) and \(f(x)\)
 one year ago

Argonx16 Group TitleBest ResponseYou've already chosen the best response.0
Sorry, my reply was written wrong. Let me restate it.
 one year ago

Argonx16 Group TitleBest ResponseYou've already chosen the best response.0
I 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))
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.3
It'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.
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.3
Look 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: \[ \frac{d}{dy}F(x) = f(y) \]
 one year ago

Argonx16 Group TitleBest ResponseYou've already chosen the best response.0
Ok, I see what you did to make F(x)=H(y)=H(g(x)) And, yes I understand the second fundamental theorem.
 one year ago

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

wio Group TitleBest ResponseYou've already chosen the best response.3
By the chain rule: \[ \frac{d}{dy}F(x) \frac{dy}{dx} = \frac{d}{dx}F(x) = f(y)\frac{dy}{dx} \]
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.3
@Argonx16 this is sort of the tricky part, but do you get it?
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.3
Another way to write this is: \[ \frac{dF(x)}{dy}\frac{dy}{dx} = \frac{dF(x)}{dx} \]Or \[ \frac{dF}{dy}\frac{dy}{dx} = \frac{dF}{dx} \]
 one year ago

Argonx16 Group TitleBest ResponseYou've already chosen the best response.0
Yes I got the chain rule step (esp after you rewrote it in a different way).
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.3
So 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: \[ f(y)\frac{dy}{dx} = f(g(x))\frac{dg(x)}{dx} = f(g(x))g'(x) \]
 one year ago

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

wio Group TitleBest ResponseYou've already chosen the best response.3
\[ \frac {d}{dx} \int\limits_{\cos 3x}^{\tan 3x} \sqrt{1+t^2} = \frac {d}{dx} \int\limits_{z}^{\tan 3x} \sqrt{1+t^2} \frac {d}{dx} \int\limits^{\cos 3x}_{a} \sqrt{1+t^2} \]
 one year ago

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

wio Group TitleBest ResponseYou've already chosen the best response.3
@Argonx16 Does it all make sense now?
 one year ago

Argonx16 Group TitleBest ResponseYou've already chosen the best response.0
Yes! @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
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.3
I've done this thing like 2 or 3 times and every time I it gets a bit easier to teach.
 one year ago

Argonx16 Group TitleBest ResponseYou've already chosen the best response.0
Yeah. It definitely helps so much more. OS is also a good way to retain math concepts taught earlier :D
 one year ago
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