# Help required: 1st and 2nd derivatives of functions

• May 18th 2009, 12:08 PM
jamm89
Help required: 1st and 2nd derivatives of functions
Im struggling with finding the 1st and 2nd derivatives of these functions (Worried)

f(x)= tan^-1 3x

So far i have done:
dy/dx = 3 (1/1+(3x)^2) = 3/ (1+9x^2)

Im not sure where to go from there
Any help much appraciated.
• May 18th 2009, 12:17 PM
Jester
Quote:

Originally Posted by jamm89
Im struggling with finding the 1st and 2nd derivatives of these functions (Worried)

f(x)= tan^-1 3x

So far i have done:
dy/dx = 3 (1/1+(3x)^2) = 3/ (1+9x^2)

Im not sure where to go from there
Any help much appraciated.

Use the chain rule on $\displaystyle 3(1+9x^2)^{-1}$ to find $\displaystyle \frac{d^2y}{dx^2}$
• May 18th 2009, 12:33 PM
jamm89
$\displaystyle (3+27x^2)^{-1}$ so i use chain rule on this?

u=$\displaystyle 3+27x^2$ y=$\displaystyle u^-1$
du/dx = 54x dy/du= $\displaystyle -u^-2$

Can someone finish this off for me please as i have no idea where to go from here
thanks
• May 18th 2009, 12:44 PM
Jester
Quote:

Originally Posted by jamm89
$\displaystyle (3+27x^2)^{-1}$ so i use chain rule on this?

u=$\displaystyle 3+27x^2$ y=$\displaystyle u^-1$
du/dx = 54x dy/du= $\displaystyle -u^-2$

Can someone finish this off for me please as i have no idea where to go from here
thanks

$\displaystyle y'' = -3(1 + 9x^2)^{-2} 18x$