Help required: 1st and 2nd derivatives of functions

**jamm89** · May 18th 2009, 12:08 PM

Im struggling with finding the 1st and 2nd derivatives of these functions

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.

**Danny** · May 18th 2009, 12:17 PM

Originally Posted by jamm89

Im struggling with finding the 1st and 2nd derivatives of these functions

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 $3(1+9x^2)^{-1}$ to find $\frac{d^2y}{dx^2}$

**jamm89** · May 18th 2009, 12:33 PM

$(3+27x^2)^{-1}$ so i use chain rule on this?

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

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

**Danny** · May 18th 2009, 12:44 PM

Originally Posted by jamm89

$(3+27x^2)^{-1}$ so i use chain rule on this?

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

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

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

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