## Mertsj 4 years ago Derivative question

1. myininaya

yes?

2. Mertsj

|dw:1328147554735:dw|

3. anonymous

ho ho ho

4. Mertsj

Hi Santa

5. myininaya

Let F be the antiderivative of $f(x)=\frac{u^2-1}{u^2+1}$ Now we have $[g(x)]'=[F(x)|_{2x}^{3x}]=[F(3x)-F(2x)]'=(3x)'f(3x)-(2x)'f(2x)$

6. myininaya

that chain rule stuff is the best! :)

7. myininaya

so we have $3 \cdot \frac{(3x)^2-1}{(3x)^2+1}-2 \cdot \frac{(2x)^2-1}{(2x)^2+1}$

8. myininaya

$3 \cdot \frac{9x^2-1}{9x^2+1}-2 \cdot \frac{4x^2-1}{4x^2+1}$

9. anonymous

$-\int _a^{2x}\frac{u^2-1}{u^2+1}du+\int_a^{3x}\frac{u^2-1}{u^2+1}du$

10. anonymous

dang did that take me a long time to type now chain rule

11. myininaya

yes it did satellite :)

12. anonymous

i am tired, time for bed. gnight

13. myininaya

me too we are going to bed

14. Mertsj

$g'(x)=\frac{-2(4x^2-1)}{4x^2+1}+\frac{3(9x^2-1)}{9x^2+1}$

15. Mertsj

Not yet, I hope.

16. Mertsj

I don't get the -2 and the 3

17. Mertsj

Maybe if I study it.

18. myininaya

:do you understand what I did?

19. Mertsj

Not quite.

20. Mertsj

I thought the u^2-1/u^2+1 would be the derivative and all I had to do was plug in the interval endpoints.

21. myininaya

$g(x)=\int\limits_{2x}^{3x}f(u) du=F(u)|_{2x}^{3x}=F(3x)-F(2x)$ Do you understand this part?

22. Mertsj

Why isn't f(u) the g'(x)?

23. myininaya

no $g(x)=F(3x)-F(2x)$

24. myininaya

g is the integral from 2x to 3x of f(u) with respect to u

25. myininaya

now since we know that $g(x)=F(3x)-F(2x)$ we can differentiate both sides

26. Mertsj

Ok. I get that and I understand what you wrote about F(3x)-F(2x)

27. myininaya

$g'(x)=(3x)'f(3x)-(2x)'f(2x)$

28. myininaya

i used chain rule for this

29. Mertsj

One second.

30. myininaya

$g'(x)=3 f(3x)-2 f(2x)$ since (3x)'=3 and (2x)'=2

31. myininaya

ok one sec granted

32. Mertsj

so F(3x)=(3x)'f(x)?

33. Mertsj

And what happened to F(x)?

34. myininaya

well (F(3x))'=(3x)'f(3x) by chain rule

35. myininaya

remember we differentiate both sides so we could find g'

36. Mertsj

Yep. Ok. I think i can figure it out now. thank you very much and how did you get so smart?

37. myininaya

i'm a little smart not much

38. myininaya

well i'm not really smart lol

39. myininaya

but thanks mert

40. Mertsj

thanks to you and sleep well...

41. myininaya

thanks i'm going to sleep for reals i'm so tired by the way satellite likes to do these types of problems another way but i love my way

42. myininaya

he likes to split the integral at a constant

43. myininaya

and then use the other part of the fundamental thm of calculus

44. myininaya

goodnight! :)

45. Mertsj

Satellite is very smart also. Do you know him personally?