# find the second derivative of...

• Feb 16th 2009, 06:23 PM
obsmith08
find the second derivative of...
Find the second derivative of f(x)= sq rt(x^2 + 1)

I think the first derivative is x(x^2 + 1)^-1/2

But, I don't know how to find the derivative of that?

A) x/sq rt (x^2 + 1)

B) sq rt(x^2 + 1) - 1/2(x^2 + 1)^-1/2

C) 2x(sq rt(x^2 + 1) + 1/2(x^2 + 1)^-1/2

D) 1/sq rt((x^2 + 1)^3)
• Feb 16th 2009, 07:06 PM
Chris L T521
Quote:

Originally Posted by obsmith08
Find the second derivative of f(x)= sq rt(x^2 + 1)

I think the first derivative is x(x^2 + 1)^-1/2

But, I don't know how to find the derivative of that?

A) x/sq rt (x^2 + 1)

B) sq rt(x^2 + 1) - 1/2(x^2 + 1)^-1/2

C) 2x(sq rt(x^2 + 1) + 1/2(x^2 + 1)^-1/2

D) 1/sq rt((x^2 + 1)^3)

Now apply the product rule, which states $\displaystyle \frac{\,d}{\,dx}\left[f\!\left(x\right)g\!\left(x\right)\right]=f\!\left(x\right)\frac{\,dg}{\,dx}+g\!\left(x\rig ht)\frac{\,df}{\,dx}$. Treat $\displaystyle x$ as $\displaystyle f\!\left(x\right)$ and $\displaystyle \left(x^2+1\right)^{-1}$ as $\displaystyle g\!\left(x\right)$.