# Thread: Second derivative

1. ## Second derivative

$\displaystyle g(t) =-\frac{4}{(t+2)^2}$

Need some assistance finding the second derivartive of this equation. If you can show me in a step by step fashion how to solve this that'd be great.

2. $\displaystyle g(t)=-4(t+2)^{-2}$
$\displaystyle g'(t)=-4(-2)(t+2)^{-3}=8(t+2)^{-3}$
$\displaystyle g''(t)=8(-3)(t+2)^{-4}=-24(t+2)^{-4}=\ -\frac{24}{(t+2)^4}$

3. Originally Posted by hemi
$\displaystyle g(t) =-\frac{4}{(t+2)^2}$

Need some assistance finding the second derivartive of this equation. If you can show me in a step by step fashion how to solve this that'd be great.

Sure buddy.

$\displaystyle \frac{d}{dt}[-\frac{4}{(t+2)^2}]=\frac{d}{dt}[-4(t+2)^{-2}]$

Now by the product and chain rules, we defferentiate once to get:

$\displaystyle \frac{d}{dt}[-4(t+2)^{-2}]=-4(-2)(t+2)^{-3}(1)+0(t+2)^{-2}=8(t+2)^{-3}$

Now to find the second derivative, we just defferentiate again:

$\displaystyle \frac{d^2}{dt^2}[-\frac{4}{(t+2)^2}]=\frac{d}{dt}[8(t+2)^{-3}]=8(-3)(t+2)^{-4}(1)+0(t+2)^{-3}=-24(t+2)^{-4}$

therefore $\displaystyle \frac{d^2g}{dt^2}=\frac{-24}{(t+2)^4}$

and there you have it. any questions?

The product rule was unnecessary here. I only did it to illustrate that the thing would stillbe valid.