# another question

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• Dec 13th 2008, 07:40 PM
mech.engineer.major
another question
I have another question if someone wouldn't mind helping me.

how would you differentiate:

y = ((x^2+4)^2) * ((2x^3-1)^3)

My first thought was that it involved two separate Chain Rules within a Product Rule. This seemed to work until I got a little stuck on how to simplify.
• Dec 13th 2008, 07:44 PM
mr fantastic
Quote:

Originally Posted by mech.engineer.major
I have another question if someone wouldn't mind helping me.

how would you differentiate:

y = ((x^2+4)^2) * ((2x^3-1)^3)

My first thought was that it involved two separate Chain Rules within a Product Rule. This seemed to work until I got a little stuck on how to simplify.

Your first thought is correct. Please show your working and where you get stuck.
• Dec 13th 2008, 08:20 PM
mech.engineer.major
this is how far I got:

y= ((x^2+4)^2) * ((2x^3-1)^3)

y= (((x^2+4)^2) * (3(2x^3-1)^2)*(6x^2)) + (((2x^3-1)^3) * (2(x^2+4)2x)

y= (((x^2+4)^2) * (18x^2(2x^3-1)^2)) + (((2x^3-1)^3) * (4x^3+16x))

y= ?

As far as I can tell I performed the Product Rule with the Chain Rules correctly but I'm not sure how it wraps up.
• Dec 13th 2008, 08:48 PM
mr fantastic
Quote:

Originally Posted by mech.engineer.major
this is how far I got:

y= ((x^2+4)^2) * ((2x^3-1)^3)

dy/dx = (((x^2+4)^2) * (3(2x^3-1)^2)*(6x^2)) + (((2x^3-1)^3) * (2(x^2+4)2x)

dy/dx = (((x^2+4)^2) * (18x^2(2x^3-1)^2)) + (((2x^3-1)^3) * (4x^3+16x))

dy/dx = ?

As far as I can tell I performed the Product Rule with the Chain Rules correctly but I'm not sure how it wraps up.

Please note the correction of your notation (in red).

Now you factorise the expression by taking out the common factor, which appears to be $2x (2x^3 - 1)^2 (x^2 + 4)$ ....
• Dec 13th 2008, 09:34 PM
mech.engineer.major
I must have blanked on factoring, it may have been hard to see for me because of the large scale equation. I worked the rest of it out and checked my answer, everything's right. Thanks