Thread: Finding the second derivative of a function

1. Finding the second derivative of a function

$\displaystyle y = (x^3 - 2x)^4$

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

$\displaystyle y'' = ?$

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

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

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

$\displaystyle y'' = ?$

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

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

3. Originally Posted by Prove It
Use the product rule.

$\displaystyle y'' = 4\left\{(x^3 - 2x)^3 \frac{d}{dx}(3x^2 - 2) + (3x^2 - 2)\frac{d}{dx}[(x^3 - 2x)^3]\right\}$
Hi again, using the product rule I get;

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

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

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

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

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

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

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

Now take out common factors.

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

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

$\displaystyle = 12(x^3 - 2x)^2(11x^4 - 16x^2 + 4)$

5. Originally Posted by Prove It

$\displaystyle = 12(x^3 - 2x)^2[2x(x^3 - 2x) + (3x^2 - 2)^2]$
What happened to the 6x?

6. Originally Posted by Archduke01
What happened to the 6x?
I took out a common factor of 3, which leaves the $\displaystyle 2x$.

7. Originally Posted by Prove It
I took out a common factor of 3, which leaves the $\displaystyle 2x$.
The common factor of 3 was only common with $\displaystyle 6x$? Why did you take out a factor 3 specifically?

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

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

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

Therefore I can take out 3 as a common factor.

9. Originally Posted by Prove It

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

You will see that the second term is $\displaystyle 3(3x - 2)^2(x^3 - 2x)^2$, 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.