Calculus - The Chain Rule

[Saint Nick]

I was just wondering if someone could explain to me how this works. I've searched Google, but I'd rather see some other people's explanation on how to do it. Here's a problem pulled from another website:

(x³+5x)^7

I just would like a straightforward, step-by-step way of doing something like this. I kind of understand how it works, but I can't get answers that match up with those in the back of the book. They always end up factoring something out and have a coefficient and some other junk... anyways, neffers and "omG go 2 G00GL3 it wil tell U" people can chime in, but any actual help is appreciated, thanks!

 

[TuxDave]

I forgot but is the chain rule something like.

d/dx of f(x)^n = n*f(x)*[d/dx of f(x)]

 

[IceBergSLiM]

 

[questro]

you have to separate the function into 2 functions. in this case, the inner and outer. So, the inner function is x^3 + 5x and the outer is x^7 and it's the outer so x= x^3 +5x right.

So, take the derivitive of the outer and get 7x where x = x^3 + 5x and the derivitive of the inner which is 3x^2 + 5. use the chain rule formula and put them together to get
7(x^3+5x) (3x^2+5) multiply that out.

the chain rule function is h'(g(x)) * g'(x) in this case h is the outer and g is the inner. See how that g(x) remains underived inside the h' thing? That's why that x^3+5x remains the same inside the 7()

 

[Kyteland]

Originally posted by: questro
you have to separate the function into 2 functions. in this case, the inner and outer. So, the inner function is x^3 + 5x and the outer is y^7 and it's the outer so y = x^3 +5x right.

So, take the derivitive of the outer and get 7y^6 where y = x^3 + 5x and the derivitive of the inner which is 3x^2 + 5. use the chain rule formula and put them together to get
7(x^3+5x)^6* (3x^2+5) multiply that out.

the chain rule function is h'(g(x)) * g'(x) in this case h is the outer and g is the inner. See how that g(x) remains underived inside the h' thing? That's why that x^3+5x remains the same inside the 7()

Fixed a few things for you. Hope I got them all!

 

[VTHodge]

Originally posted by: questro
you have to separate the function into 2 functions. in this case, the inner and outer. So, the inner function is x^3 + 5x and the outer is x^7 and it's the outer so x= x^3 +5x right.

So, take the derivitive of the outer and get 7x where x = x^3 + 5x and the derivitive of the inner which is 3x^2 + 5. use the chain rule formula and put them together to get
7(x^3+5x) (3x^2+5) multiply that out.

the chain rule function is h'(g(x)) * g'(x) in this case h is the outer and g is the inner. See how that g(x) remains underived inside the h' thing? That's why that x^3+5x remains the same inside the 7()

Good explanation. Welcome to AT.

Edit: Actually you left off the reduced exponent.

7 (x^3+5x)^6 (3x^2+5)

 

[VTHodge]

Originally posted by: Kyteland

Originally posted by: questro
you have to separate the function into 2 functions. in this case, the inner and outer. So, the inner function is x^3 + 5x and the outer is y^7 and it's the outer so y = x^3 +5x right.

So, take the derivitive of the outer and get 7y^6 where y = x^3 + 5x and the derivitive of the inner which is 3x^2 + 5. use the chain rule formula and put them together to get
7(x^3+5x)^6* (3x^2+5) multiply that out.

the chain rule function is h'(g(x)) * g'(x) in this case h is the outer and g is the inner. See how that g(x) remains underived inside the h' thing? That's why that x^3+5x remains the same inside the 7()

Fixed a few things for you. Hope I got them all!

lol . . . even better.

 

[sao123]

Originally posted by: pOwder
I was just wondering if someone could explain to me how this works. I've searched Google, but I'd rather see some other people's explanation on how to do it. Here's a problem pulled from another website:

(x³+5x)^7

I just would like a straightforward, step-by-step way of doing something like this. I kind of understand how it works, but I can't get answers that match up with those in the back of the book. They always end up factoring something out and have a coefficient and some other junk... anyways, neffers and "omG go 2 G00GL3 it wil tell U" people can chime in, but any actual help is appreciated, thanks!

d/dx (x³+5x)^7
(7 * (x³+5x)^6) * (d/dx x³+5x)
(7 * (x³+5x)^6) * (3x²+5)

 

[ArchCenturion]

now try

d/dx (d/dx (x³+5x)^7 )

 

[ArchCenturion]

d/dx (7 * (x³+5x)^6) * (3x^2+5)

I think i get

7 * 6 * (x³+5x)^5) * (3x^2+5) * (3x^2+5) + 7 * (x³+5x)^6) * 6x

42 * (x³+5x)^5) * (3x^2+5)^2 + 42x * (x³+5x)^6)

seems right to me, but im sure somebody will pipe up if its wrong...



Edit: sao123 how did you get the super scripts? I only have the ones i copy pasted.

 

[dullard]

Originally posted by: pOwder
I was just wondering if someone could explain to me how this works. I've searched Google, but I'd rather see some other people's explanation on how to do it. Here's a problem pulled from another website:

(x³+5x)^7

I just would like a straightforward, step-by-step way of doing something like this. I kind of understand how it works, but I can't get answers that match up with those in the back of the book. They always end up factoring something out and have a coefficient and some other junk... anyways, neffers and "omG go 2 G00GL3 it wil tell U" people can chime in, but any actual help is appreciated, thanks!

Just simply and do many easy steps.

1) g(x) = (x^3 + 5*x)^7. Thus the problem becomes simple to write: find g'(x).

2) f(x) = x^3 + 5*x. Thus the problem becomes: g(x)=[f(x)]^7.

You know how to take that simple derivative with the chain rule: g'(x) = 7*[f(x)]^6*f'(x).

3) Doh, we have to continue because we don't know what f'(x) is. Well look at #2 above. It should be easy to find that f'(x) = 3*x^2 + 5.

4) Plug into the answer for #2: g'(x)=7*[f(x)]^6*(3*x^2+5). You are done with the problem. Who cares what the back of the book says.

5) If you wish, you could expand it. Plug in the definition for f(x): g'(x) = 7*(x^3+5*x)^6*(3*x^2+5). Again, who cares if the back of the book did further algebra to simplify? The calculus is done, you can do algebra if you want, but you don't have to.

 

[questro]

Originally posted by: VTHodge

Originally posted by: questro
you have to separate the function into 2 functions. in this case, the inner and outer. So, the inner function is x^3 + 5x and the outer is x^7 and it's the outer so x= x^3 +5x right.

So, take the derivitive of the outer and get 7x where x = x^3 + 5x and the derivitive of the inner which is 3x^2 + 5. use the chain rule formula and put them together to get
7(x^3+5x) (3x^2+5) multiply that out.

the chain rule function is h'(g(x)) * g'(x) in this case h is the outer and g is the inner. See how that g(x) remains underived inside the h' thing? That's why that x^3+5x remains the same inside the 7()

Good explanation. Welcome to AT.

Edit: Actually you left off the reduced exponent.

7 (x^3+5x)^6 (3x^2+5)

dang!

Yea I did. Sorry

 

[dullard]

Oh and by the way, if you really, really must do more algebra the answer is:

g'(x) = 21*x^20 + 665*x^18 + 8925*x^16 + 65625*x^14 + 284375*x^12 + 721875*x^10 + 984375*x^8 + 546875*x^6

Is that what the back of the book states?

 

[Saint Nick]

I can't @$#@ing figure this out I am on the verge of tears trying to get this

I can figure out the ones that we already did in class, but the ones that are assigned I can't figure out because I don't know where to start and then what to do and what to factor out once I get going. I'm going to take a break from it for a bit and then try again in about... never.

Thanks so much so far guys, I appreciate all you've done, I've read over most of the replies so far and they all seemingly match up, I just have to find some handy way to remember this for my test next week.

 

[dullard]

Originally posted by: pOwder
I can't @$#@ing figure this out I am on the verge of tears trying to get this

I can figure out the ones that we already did in class, but the ones that are assigned I can't figure out because I don't know where to start and then what to do and what to factor out once I get going. I'm going to take a break from it for a bit and then try again in about... never.

Thanks so much so far guys, I appreciate all you've done, I've read over most of the replies so far and they all seemingly match up, I just have to find some handy way to remember this for my test next week.

You are always welcome. I don't think you can well learn by watching others do the work. Most people learn by failing and then catching the errors and doing it again without the error. So the fact that you are struggling is the first step towards learning. Don't give up on it now, because you are just now going to see results.

Instead of watching us do the problem, can you post one of the other problems you are struggling with? Then show us in detail step-by-step what you are trying. Maybe we can pinpoint your problem that you are overseeing.

 

[jman19]

Originally posted by: pOwder
I can't @$#@ing figure this out I am on the verge of tears trying to get this

I can figure out the ones that we already did in class, but the ones that are assigned I can't figure out because I don't know where to start and then what to do and what to factor out once I get going. I'm going to take a break from it for a bit and then try again in about... never.

Thanks so much so far guys, I appreciate all you've done, I've read over most of the replies so far and they all seemingly match up, I just have to find some handy way to remember this for my test next week.

Dude, it will be all right. Really you just need to remember some important rules and practice, practice, practice! If you do as many sample problems as possible before the test, you'll be fine. Just be sure to do these problems ahead of time so you can see a TA or your prof before hand. Good luck!

 

[James3shin]

outside, then inside. Take the derivative of everything on the outside of the parenthesis, in this case pretend it's (z)^7, where z = x³+5x. So far you should have 7(z)^6, now take the derivative of Z, aka dZ. dZ = (3x^2)+5. Now put the two together, the outside and inside.

 

[Saint Nick]

Here is a problem that I'm working on right now:

4y²(y²+1)^(5/4)

I found f(y), g(y), f`(y), and g`(y):

f(y) = 4y²
f`(y) = 8y
g(y) = (y² + 1)^(5/4)
g`(y) = [(5/4)(y² + 1)^(1/4)] * (2y)

So, according to (f * g') + * (f' * g)

q(y) = (4y²)[(5/4)(y² + 1)^(1/4)(2y)] + (8y)(y² + 1)^(5/4)

From there, I have no idea what to do. I don't know what to distribute where and what to factor out and where to put all these numbers! It should be easy--it's algebra. But for some reason this is hard

 

[sao123]

Originally posted by: ArchCenturion

Edit: sao123 how did you get the super scripts? I only have the ones i copy pasted.

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[dullard]

Originally posted by: pOwder
q(y) = (4y²)[(5/4)(y² + 1)^(1/4)(2y)] + (8y)(y² + 1)^(5/4)

From there, I have no idea what to do. I don't know what to distribute where and what to factor out and where to put all these numbers! It should be easy--it's algebra. But for some reason this is hard

100% correct. You are done. What is the problem? Good job.

Remember 99% of people's problems in calculus is that they can't remember algebra. You aren't alone.

 

[Alienwho]

tagged for later.

 

[oboeguy]

Originally posted by: dullard
Remember 99% of people's problems in calculus is that they can't remember algebra. You aren't alone.

So, so true.

 

[VTHodge]

I would only simplify the terms without any addition - the "monomials", i suppose.

q(y) = (4y²)[(5/4)(y² + 1)^(1/4)(2y)] + (8y)(y² + 1)^(5/4)

q(y) = (10y^3)(y² + 1)^(1/4) + (8y)(y² + 1)^(5/4)

 

[Alienwho]

I, like the OP, am nearly at the point of tears in trying to understand my calculus. Can somebody help me find the second derivative of f(x)=x^2(3x^2+3x-4)

So far i've got.

u=x^2
u'=2x
v=3x^2+3x-4
v'=6x+3

Following f'(x)=vdu+udv I have:
f'(x)=3x^2+3x-4(2x)+(x^2)(6x+3)

But now I have no idea what I can do, or how I go about finding the second derivative of it.

 

[Alienwho]

Or perhaps, somebody could explain to me the step I'm missing. My book has the following problem: f(x)=x(3x-9)^3
It lists the steps as following:

(3x-9)^3 * 1 + x(3x-9)^2 * 9 (I understand this step, but then it suddenly skips to: )
(3x-9)^2 (3x-9+9x)

How does that happen? What am I missing? Where does the (3x-9)^3 dissapear to?

I know there is just something simple I'm not getting which is throwing off everything.