## anonymous 3 years ago Calculus! (See comments)

1. anonymous

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

2. 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

3. anonymous

(Second fundamental therom)

4. anonymous

check FTC 2

5. anonymous

:)

6. anonymous

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

7. 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.

8. anonymous

Let $$y = g(x)$$

9. kirbykirby

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

10. anonymous

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

11. anonymous

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

12. kirbykirby

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

13. anonymous

See where I'm going with this?

14. anonymous

@Argonx16

15. kirbykirby

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

16. anonymous

@wio I'm reading through it (:

17. 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

18. anonymous

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

19. anonymous

@kirbykirby Ok, thanks!

20. anonymous

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

21. anonymous

@Argonx16 I'm not finished yet but sure.

22. 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

23. anonymous

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

24. anonymous

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

25. anonymous

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

26. anonymous

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

27. 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))

28. 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.

29. 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: $\frac{d}{dy}F(x) = f(y)$

30. anonymous

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

31. anonymous

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

32. anonymous

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

33. anonymous

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

34. anonymous

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

35. anonymous

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

36. 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: $f(y)\frac{dy}{dx} = f(g(x))\frac{dg(x)}{dx} = f(g(x))g'(x)$

37. anonymous

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

38. anonymous

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

39. anonymous

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

40. anonymous

@Argonx16 Does it all make sense now?

41. 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

42. anonymous

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

43. anonymous

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