Calculus! (See comments)

- anonymous

Calculus! (See comments)

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

\[\frac {d}{dx} \int\limits_{\cos 3x}^{\tan 3x} \sqrt{1+t^2}\] Preferably with steps+rationale shown. Thanks!

- anonymous

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

- anonymous

(Second fundamental therom)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

check FTC 2

- anonymous

:)

- anonymous

Suppose you have the following: \[ F(x) = \int_a^{g(x)}f(t)dt \]

- anonymous

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.

- anonymous

Let \(y = g(x)\)

- kirbykirby

You would have to split the integral at some constant (say a)

- anonymous

Then suppose : \[ H(y) = \int_a^yf(t)dt \]Notice that \(H(y) = H(g(x)) = F(x)\)

- anonymous

By the fundamental theorem: \[ \frac{d}{dy}H(y) = f(y) = f(g(x)) \]

- kirbykirby

\[\frac {d}{dx} \int\limits\limits_{\cos 3x}^{\tan 3x} \sqrt{1+t^2}dt = \frac {d}{dx} \int\limits\limits_{\cos 3x}^{a} \sqrt{1+t^2}dt+\frac {d}{dx} \int\limits\limits_{a}^{\tan 3x} \sqrt{1+t^2}dt\]

- anonymous

See where I'm going with this?

- anonymous

@Argonx16

- kirbykirby

and \[\int\limits\limits_{\cos 3x}^{a} \sqrt{1+t^2}=-\int\limits\limits_{a}^{\cos3x} \sqrt{1+t^2}\]

- anonymous

@wio I'm reading through it (:

- kirbykirby

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

- anonymous

Basically we have: \[ \frac{d}{dy}F(x) = f(g(x)) \]

- anonymous

@kirbykirby Ok, thanks!

- anonymous

@wio here, let me put what you said in words to see if I get it

- anonymous

@Argonx16 I'm not finished yet but sure.

- anonymous

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

- anonymous

When we say \(y=g(x)\) we're just saying they are interchangeable.

- anonymous

The reason I did that was to make it so you can see that a function is sort of just like a variable.

- anonymous

also there is a difference between \(F(x)\) and \(f(x)\)

- anonymous

Sorry, my reply was written wrong. Let me restate it.

- anonymous

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))

- anonymous

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.

- anonymous

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: \[ \frac{d}{dy}F(x) = f(y) \]

- anonymous

Ok, I see what you did to make F(x)=H(y)=H(g(x)) And, yes I understand the second fundamental theorem.

- anonymous

Now the trick here is to multiply both sides by \(dy/dx\) \[ \frac{d}{dy}F(x) \frac{dy}{dx} = f(y)\frac{dy}{dx} \]

- anonymous

By the chain rule: \[ \frac{d}{dy}F(x) \frac{dy}{dx} = \frac{d}{dx}F(x) = f(y)\frac{dy}{dx} \]

- anonymous

@Argonx16 this is sort of the tricky part, but do you get it?

- anonymous

Another way to write this is: \[ \frac{dF(x)}{dy}\frac{dy}{dx} = \frac{dF(x)}{dx} \]Or \[ \frac{dF}{dy}\frac{dy}{dx} = \frac{dF}{dx} \]

- anonymous

Yes I got the chain rule step (esp after you rewrote it in a different way).

- anonymous

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: \[ f(y)\frac{dy}{dx} = f(g(x))\frac{dg(x)}{dx} = f(g(x))g'(x) \]

- anonymous

Putting it all together: \[ \frac{d}{dx} \left[ \int\limits_{a}^{g(x)}f(t)dt \right] = f(g(x))g′(x) \]

- anonymous

\[ \frac {d}{dx} \int\limits_{\cos 3x}^{\tan 3x} \sqrt{1+t^2} = \frac {d}{dx} \int\limits_{z}^{\tan 3x} \sqrt{1+t^2} -\frac {d}{dx} \int\limits^{\cos 3x}_{a} \sqrt{1+t^2} \]

- anonymous

To expand on the earlier realization: \[ \frac{d}{dx} \left[ \ \int\limits_{h(x)}^{g(x)}f(t)dt \right] = f(g(x))g′(x) -f(h(x))h'(x) \]

- anonymous

@Argonx16 Does it all make sense now?

- anonymous

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

- anonymous

I've done this thing like 2 or 3 times and every time I it gets a bit easier to teach.

- anonymous

Yeah. It definitely helps so much more. OS is also a good way to retain math concepts taught earlier :D

Looking for something else?

Not the answer you are looking for? Search for more explanations.