# Chain rule

1. Mar 16, 2014

### anthonyk2013

Differentiate the following by rule y=(2x2+4x)5

Is the chain rule the right rule to use?

dy/dx=dy/du*du/dx

Let U=2x2+4x

du/dx=4x+4

y=(u)5 → dy/du=5(u)4

dy/dx=5(u)4*4x+4

dy/dx=5(2x2+4x)4*4x+4

dy/dx= 30(2x2+4x)44x

dy/dx= 30(2x216x)4

I'm wondering if I am on the right track?

2. Mar 16, 2014

### Staff: Mentor

Anthonyk, your questions should be posted in the Homework & Coursework sections (Calculus & Beyond) - not in the technical math sections.

3. Mar 16, 2014

### anthonyk2013

ok no problem

4. Mar 16, 2014

### Ray Vickson

I hope you did not mean what you wrote, which was
$$\frac{dy}{dx} = 5(x^2 + 4x)^4 4x + 4, \text{ which } = 4 +5(x^2 + 4x)^4 4x$$
I hope you meant
$$\frac{dy}{dx} = 5(x^2 + 4x)^4 (4x + 4)$$
If that is what you did mean, that is what you should write; note the parentheses.

5. Mar 16, 2014

### Staff: Mentor

Use parentheses where they are needed.
The right side should be 5u4 * (4x + 4)
That last factor should be (4x + 4)
No. I can't tell what you did here. How did you get 30 at the beginning of the right side?

6. Mar 16, 2014

### vela

Staff Emeritus
In addition to what Ray noted, I can't figure out what you did to get the last two lines. You need to go back and review algebra.

7. Mar 16, 2014

### anthonyk2013

Ya sorry that is what I meant.

can I simplify this further or can I leave it like that?

8. Mar 16, 2014

### Staff: Mentor

You can factor 4 out of the 4x + 4 term, and put it with the 5 factor. Otherwise, that's about all you can do. For most purposes, leaving it in factored form is preferable to multiplying everything out.

9. Mar 16, 2014

### anthonyk2013

dy/dx=20(2x2+4x)*(4x) this what you mean

10. Mar 16, 2014

### Staff: Mentor

No. Like vela said, you need to take some time to review algebra.

Starting from here:
dy/dx=5(2x2+4x)4 * (4x+4), factor 4 out of the last expression in parentheses, and combine that 4 with the leading 5. You did this, but the problem is that 4x + 4 ≠ 4*x. That seems to be what you're doing.

11. Mar 17, 2014

### anthonyk2013

dy/dx=5+4(2x2+4x)44x

dy/dx=9(2x2+4x)44x

or

dy/dx=10x2+20x*(4x+4)

Last edited: Mar 17, 2014
12. Mar 17, 2014

### BruceW

none of those. take your time, just using rules of arithmetic that you are certain about. for example, (4x+4) = 4*(x+1) right? So then what does the equation look like?

13. Mar 17, 2014

### anthonyk2013

I know this is probably very simple I just cant get it.

dy/dx=5(2x2+4x)*(4x+4)

Do I separate it out and treat 5(2x2+4x) from (4x+4)

5(2x2+4x)*4(x+1)?

14. Mar 17, 2014

### BruceW

yeah, almost. you forgot the first bit should be to the power of four. so it is 5(2x2+4x)4*4(x+1) And yes, it is fine to 'separate out'. In arithmetic, it is always OK to say a*(b*c)=a*b*c i.e. in this case 5(2x2+4x)4*(4x+4) = 5(2x2+4x)4*4(x+1)

15. Mar 17, 2014

### anthonyk2013

5(2x2+4x)4*4(x+1)

Thanks very much, frustrating that its so simple. thanks again

16. Mar 17, 2014

### BruceW

glad to have helped! yeah, I'm writing Makefiles at the moment, which should be a simple programming thing to do. But it's taking me ages! haha

17. Mar 17, 2014

### anthonyk2013

Best of luck.

18. Mar 17, 2014

### Staff: Mentor

You are really not going to be able to do calculus unless you have good facility with algebra. You did not realize that you could combine the factors 5 and 4 to give 20. Your previous posts had several algebra errors in them. I implore you to please follow Vela's advice and review algebra.

Chet

19. Mar 17, 2014

### anthonyk2013

Thanks I have been looking over it today. It's just the work load of study, job and kids I can only do so much.

I can multiply 5*4 to get 20. Can I do anything with what in the bracket?

20. Mar 17, 2014

### Staff: Mentor

Yes, you can factor out a 2, and, when it comes out of the bracket, it becomes 24=16, which, when multiplied by the 20 becomes 320.

Chet