## Study23 2 years ago {Derivatives, Focus: Chain Rule. Help appreciated} $$\ \Huge y=cos(a^3+x^3) .$$

1. Dido525

Is a a constant?

2. campbell_st

are you differentiating with respect to x..?

3. Dido525

or a? :P .

4. Study23

Yes @campbell_st. I'm not sure @Dido525

5. Dido525

That means a is a constant I assume O_o .

6. Study23

$$\ \text{ I got to this point, but my text tells me this answer is incorrect: } y'=-sinx(3a^2+3x^2).$$

7. Dido525

|dw:1352349392745:dw|

8. campbell_st

let u = a^3 + x^3 du/dx = 3x^2 let y = cos(u) du/dy = -sin (u) so dy/dx = dy/du * du/dx

9. Dido525

I can't see the rest.

10. Study23

One moment

11. Study23

$$\ \Huge -sinx(3a^2+3x^2)$$

12. campbell_st

then replace u with a^3 + x^3 for the final solution.

13. campbell_st

nope its dy/dx = -sin(a^3 + x^3) * 3x^2

14. Study23

So the way my math teacher did this in class was using @Dido525 illustrated

15. Study23

@campbell_st How did you get that?

16. Dido525

Because a is a constant. The derivative of a constant is 0.

17. Dido525

We multiply by the derivative of the inside part when using the chain rule.

18. campbell_st

and I used the chain rule..

19. aerokat25

@camp is correct per calculus instructor

20. campbell_st

the a^3 x^3 doesn't change

21. Study23

@campbell_st I thought a was a constant then? How does that stay the same?

22. Dido525

Here: |dw:1352349668500:dw|

23. Dido525

We DO NOT change the inside.

24. aerokat25

Now if you are implicitly differentiating then you could possibly get a different answer if a is not constant. However I'm pretty sure a is a constant in this case

25. Study23

|dw:1352349758941:dw|

26. Dido525

Differentiate that.

27. Dido525

With respect to x. Meaning a is a constant. We know the derivative of a constant is 0.

28. thewilled1

Usually letters a, b, and c are considered constants.

29. Study23

So you can't differentiate more than one variable in any given function? Only variable is differentiated in a problem with more than one variable?

30. aerokat25

Also check to see if you typed the question correctly because that answer you typed from the back of the book looks like you changed the angle measurement to x

31. Dido525

Well there are partial derivatives but you probably won't learn that now.

32. Study23

@aerokat25 The correct answer the book provides is $$\ y'=-3x^2sin(a^3+x^3) .$$ Why can't I differentiate x^3 ?

33. campbell_st

its just like (x^2 + 3)^4 dy/dx = 4(x^2 + 3) ^3 *2x the brackets don't change

34. campbell_st

you do... thats how you get 3x^2

35. Dido525

When you have a function Just inside another function the function inside the brackets DO NOT change.

36. Study23

But the term is still inside the parenthesis? That's really confusing me right now

37. Study23

parentheses*

38. campbell_st

that is why textbooks talk about u = a^3 + x^3 du/dx = 3x^2 y = cos(u) dy/du = -sin(u) dy/dx = dy/du * du/dx

39. Dido525

|dw:1352350103544:dw|

40. aerokat25

Chain rule says d/dx(cos u)= -sin u (u')

41. campbell_st

yep and du/dx = u' you are differentiating u with respect to x

42. Study23

Hmmm. That's really confusing because our teacher shows us the u method on one example, and said he doesn't use it. So I don't really follow, to be honest.

43. aerokat25

So u= a^3 + x^3

44. aerokat25

U'= 0 + 3x^2

45. campbell_st

well its the derivative of the outer function cos times the derivative of the inner function a^3 + x^3

46. Study23

Okay, I get that part, but the one thing I am not understanding is how the expression remains. I'm sorry if I'm being silly or something.

47. aerokat25

Ok well u know d/dx of cos x= -sin x right?

48. Dido525

The expression inside of the expression on the outside ALWAYS remains. It's just how it is.

49. Study23

$$\ \text{Yes!}$$

50. Study23

"on the outside"?

51. aerokat25

Well d/dx cos x actually equals -sin x (x') but the derivative of x is just 1 so you have actually been using chain all along and didn't really know it.

52. Study23

Okay$$\ \text{...}$$

53. aerokat25

And that angle measure x has never changed right? So you aren't going to change what you're given as an angle measurement now either. You just have to multiply by its derivative which isn't going to be a 1 anymore

54. Study23

Angle measure?