Calculus - The Chain Rule

[blueshoe]

Originally posted by: 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.

They factor out a (3x-9)^2 term from (3x-9)^3 * 1 + x(3x-9)^2 * 9 .

 

[Alienwho]

Originally posted by: blueshoe

Originally posted by: 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.

They factor out a (3x-9)^2 term from (3x-9)^3 * 1 + x(3x-9)^2 * 9 .

Whoa that's right, I see it now. Thanks!

 

[cliftonite]

Here is a good site that has flash based explanations. Try to work their problems out step by step and see how they do it.

 

[SpecialEd]

Originally posted by: 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.

first multiply the function out to f(x)=3x^4+3x^3-4x^2. Then take the derivative. Don't make life harder than it has to be.

 

[blueshoe]

Originally posted by: 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.

For this problem you used the product rule to differentiate, but since it is pretty simple I'd just multyply out first to get f(x)=3x^4+3x^3-4x^2. Differentiating that is a lot easier IMO. You will end up with a cleaner f'(x) as well.

But your f'(x) equation is still correct. It just isn't really convinient to find the derivative of that. Once again, just multiply it out and you should get an equation with 3 terms in it.

Edit: ok I didnt see SpecialEd's post

 

[Alienwho]

Jeez you guys are right, i'm an idiot. That's half my problem, I don't know when to simply multiply it out, or factor it out, or when to use the product rule, or chain rule, or anything else for that matter.

 

[Saint Nick]

Originally posted by: 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)

Okay, I made it that far just a bit ago... and ended up with this:

q'(y) = (80y^4)(y² + 1)^(1/4)(y² + 1)^(5/4)

The answer in the back of the book got this answer:

q'(y) = 2y(y² + 1)^(1/4)(9y²+ + 4)

any help there?