# Finding the second derivative of a function

• Nov 13th 2009, 07:34 PM
Archduke01
Finding the second derivative of a function
$y = (x^3 - 2x)^4$

$y' = 4(x^3 - 2x)^3 (3x^2 - 2)$

$y'' = ?$

I'm having difficulty proceeding from there to get the second derivative. The answer is $12(x^3 - 2x)^2 (11x^4 - 16x^2 + 4)$
• Nov 13th 2009, 07:37 PM
Prove It
Quote:

Originally Posted by Archduke01
$y = (x^3 - 2x)^4$

$y' = 4(x^3 - 2x)^3 (3x^2 - 2)$

$y'' = ?$

I'm having difficulty proceeding from there to get the second derivative. The answer is $12(x^3 - 2x)^2 (11x^4 - 16x^2 + 4)$

Use the product rule.

$y'' = 4\left\{(x^3 - 2x)^3 \frac{d}{dx}(3x^2 - 2) + (3x^2 - 2)\frac{d}{dx}[(x^3 - 2x)^3]\right\}$
• Nov 13th 2009, 07:44 PM
Archduke01
Quote:

Originally Posted by Prove It
Use the product rule.

$y'' = 4\left\{(x^3 - 2x)^3 \frac{d}{dx}(3x^2 - 2) + (3x^2 - 2)\frac{d}{dx}[(x^3 - 2x)^3]\right\}$

Hi again, using the product rule I get;

$y'' = 4[3(x^3 - 2x)^2 (3x^2 - 2) (3x^2 - 2) + (x^3 - 2x)^3 (6x)$

I'm not sure if it's correct or not but regardless I don't know how to simplify the above expression.
• Nov 13th 2009, 07:56 PM
Prove It
Quote:

Originally Posted by Archduke01
Hi again, using the product rule I get;

$y'' = 4[3(x^3 - 2x)^2 (3x^2 - 2) (3x^2 - 2) + (x^3 - 2x)^3 (6x)$

I'm not sure if it's correct or not but regardless I don't know how to simplify the above expression.

Yes, now you can take out common factors

$4\left\{(x^3 - 2x)^3 \frac{d}{dx}(3x^2 - 2) + (3x^2 - 2)\frac{d}{dx}[(x^3 - 2x)^3]\right\}$

$= 4[6x(x^3 - 2x)^3 + 3(3x^2 - 2)(3x^2 - 2)(x^3 - 2x)^2]$

Now take out common factors.

$= 12(x^3 - 2x)^2[2x(x^3 - 2x) + (3x^2 - 2)^2]$

$= 12(x^3 - 2x)^2[2x^4 - 4x^2 + 9x^4 - 12x^2 + 4]$

$= 12(x^3 - 2x)^2(11x^4 - 16x^2 + 4)$
• Nov 13th 2009, 08:08 PM
Archduke01
Quote:

Originally Posted by Prove It

$= 12(x^3 - 2x)^2[2x(x^3 - 2x) + (3x^2 - 2)^2]$

What happened to the 6x?
• Nov 13th 2009, 08:10 PM
Prove It
Quote:

Originally Posted by Archduke01
What happened to the 6x?

I took out a common factor of 3, which leaves the $2x$.
• Nov 13th 2009, 08:12 PM
Archduke01
Quote:

Originally Posted by Prove It
I took out a common factor of 3, which leaves the $2x$.

The common factor of 3 was only common with $6x$? Why did you take out a factor 3 specifically?
• Nov 13th 2009, 08:15 PM
Prove It
Quote:

Originally Posted by Archduke01
The common factor of 3 was only common with $6x$? Why did you take out a factor 3 specifically?

The first term is $6x(x^3 - 2x)^3$ has a factor of 3.

You will see that the second term is $3(3x - 2)^2(x^3 - 2x)^2$, which also has a factor of 3.

Therefore I can take out 3 as a common factor.
• Nov 13th 2009, 08:17 PM
Archduke01
Quote:

Originally Posted by Prove It

The first term is $6x(x^3 - 2x)^3$ has a factor of 3.

You will see that the second term is $3(3x - 2)^2(x^3 - 2x)^2$, which also has a factor of 3.

Therefore I can take out 3 as a common factor.

Oh right, I see it. I missed it originally because I had already taken out the 3 in the second term by accident.

Thanks for your help; I understand it now. And sorry for taking up your time with questions.