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

1. Dido525 Group Title

Is a a constant?

2. campbell_st Group Title

are you differentiating with respect to x..?

3. Dido525 Group Title

or a? :P .

4. Study23 Group Title

Yes @campbell_st. I'm not sure @Dido525

5. Dido525 Group Title

That means a is a constant I assume O_o .

6. Study23 Group Title

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

7. Dido525 Group Title

|dw:1352349392745:dw|

8. campbell_st Group Title

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 Group Title

I can't see the rest.

10. Study23 Group Title

One moment

11. Study23 Group Title

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

12. campbell_st Group Title

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

13. campbell_st Group Title

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

14. Study23 Group Title

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

15. Study23 Group Title

@campbell_st How did you get that?

16. Dido525 Group Title

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

17. Dido525 Group Title

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

18. campbell_st Group Title

and I used the chain rule..

19. aerokat25 Group Title

@camp is correct per calculus instructor

20. campbell_st Group Title

the a^3 x^3 doesn't change

21. Study23 Group Title

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

22. Dido525 Group Title

Here: |dw:1352349668500:dw|

23. Dido525 Group Title

We DO NOT change the inside.

24. aerokat25 Group Title

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 Group Title

|dw:1352349758941:dw|

26. Dido525 Group Title

Differentiate that.

27. Dido525 Group Title

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

28. thewilled1 Group Title

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

29. Study23 Group Title

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 Group Title

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 Group Title

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

32. Study23 Group Title

@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 Group Title

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

34. campbell_st Group Title

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

35. Dido525 Group Title

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

36. Study23 Group Title

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

37. Study23 Group Title

parentheses*

38. campbell_st Group Title

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 Group Title

|dw:1352350103544:dw|

40. aerokat25 Group Title

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

41. campbell_st Group Title

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

42. Study23 Group Title

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 Group Title

So u= a^3 + x^3

44. aerokat25 Group Title

U'= 0 + 3x^2

45. campbell_st Group Title

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

46. Study23 Group Title

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 Group Title

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

48. Dido525 Group Title

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

49. Study23 Group Title

$$\ \text{Yes!}$$

50. Study23 Group Title

"on the outside"?

51. aerokat25 Group Title

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 Group Title

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

53. aerokat25 Group Title

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 Group Title

Angle measure?