# Thread: chain rule second derivative

1. ## chain rule second derivative

I'm supposed to find the second derivative of $\displaystyle (x^2+5)^8$
What I got was: $\displaystyle f(u)=u^8, u(x)=(x^2+5)$
$\displaystyle (8(x^2+5)^7)(2)=16(x^2+5)^7$
$\displaystyle f(u)=16(u)^7, u(x)=(x^2+5)$
$\displaystyle (16)7(x^2+5)^6(2)=224(x^2+5)^6$
but the back of the book says it's: $\displaystyle 80(x^2+5)^6(3x^2+1)$
Any help would be much appreciated!

2. $\displaystyle f = (x^2 + 5)^8$.

Let $\displaystyle u = x^2 + 5$ so that $\displaystyle f = u^8$.

$\displaystyle \frac{du}{dx} = 2x$.

$\displaystyle \frac{df}{du} = 8u^7 = 8(x^2 + 5)^7$.

Therefore $\displaystyle \frac{df}{dx} = 16x(x^2 + 5)^7$.

Now you will need to use the product rule...

$\displaystyle \frac{d^2f}{dx^2} = 16x\cdot 2x \cdot 7(x^2 + 5)^6 + 16(x^2 + 5)^7$

$\displaystyle = 224x^2(x^2 + 5)^6 + 16(x^2 + 5)^7$

$\displaystyle = 16(x^2 + 5)^6(14x^2 + x^2 + 5)$

$\displaystyle = 16(x^2 + 5)^6(15x^2 + 5)$

$\displaystyle = 80(x^2 + 5)^6(3x^2 + 1)$.

3. oh duh, 2x not 2. thanks!