Applying the Chain Rule to y=(2x2+4x)5

[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?

 

[Mark44]

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

 

[anthonyk2013]

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

ok no problem

 

[Ray Vickson]

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?

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

 

[Mark44]

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

Use parentheses where they are needed.
The right side should be 5u⁴ * (4x + 4)

dy/dx=5(2x²+4x)⁴*4x+4

That last factor should be (4x + 4)

dy/dx= 30(2x²+4x)⁴4x

No. I can't tell what you did here. How did you get 30 at the beginning of the right side?

dy/dx= 30(2x²16x)⁴

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

 

[vela]

dy/dx=5(2x²+4x)⁴*4x+4

dy/dx= 30(2x²+4x)⁴4x

dy/dx= 30(2x²16x)⁴

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.

 

[anthonyk2013]

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

Ya sorry that is what I meant.

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

 

[Mark44]

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

Ya sorry that is what I meant.

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

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.

 

[anthonyk2013]

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.

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

 

[Mark44]

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

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.

 

[anthonyk2013]

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.

dy/dx=5+4(2x²+4x)⁴4x

dy/dx=9(2x²+4x)⁴4x

or

dy/dx=10x²+20x*(4x+4)

 

[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?

 

[anthonyk2013]

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?

I know this is probably very simple I just can't 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)?

 

[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)

 

[anthonyk2013]

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 abc=a(bc) i.e. in this case 5(2x2+4x)⁴*(4x+4) = 5(2x²+4x)⁴*4(x+1)

5(2x²+4x)⁴*4(x+1)

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

 

[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

 

[anthonyk2013]

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

Best of luck.

 

[Chestermiller]

Best of luck.

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

 

[anthonyk2013]

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

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?

 

[Chestermiller]

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?

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

 

[vela]

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.

You'll be much more efficient at studying once you get the algebra down. You don't want to be wasting time struggling with algebra instead of concentrating on the material you're supposed to be learning. I'll note that with this problem, you had the calculus part — applying the chain rule — done correctly; it was only the subsequent algebra that was a problem. It might feel like reviewing algebra is a low priority, but you really should make it a higher one.

 

[Mark44]

I agree with what vela and Chet have said. I recognize that you have a lot of time constraints at the moment, with kids, job, and school, but a small amount of time spent at bolstering your algebra skills (say, 30 minutes of focused effort a day) will save a lot of time down the road.

 

[anthonyk2013]

I will try and put some time aside for algebra, thanks for advice and help I appreciate it. Cheers and happy st Patrick's day :-)