# Thread: first and second derivatives of the function

1. ## first and second derivatives of the function

Find both the first and second derivatives of the the function:
$\displaystyle f(x) = tan^-1 3x$
I started by doing
$\displaystyle dy/dx = 3* 1/1+(3x)^2$
Then $\displaystyle 3/(1+9x^2)$
Can anyone finish this off for me please.

2. $\displaystyle f(x)=tan^-1(u(x))$

$\displaystyle f '(x)=\frac{u'(x)}{1+u^2(x)}$

$\displaystyle f''(x)=\frac{(1+u^2)u''(x) - (u'(x))(2u'(x)u(x))}{(1+u^2(x))^2}$

3. Another way to get the second derivative is to write
[/tex]f'(x)= 3(1+ 9x^2)^{-1}[/tex] and use the power rule and chain rule.

4. $\displaystyle f(x)=tan^-1(3(x))$

$\displaystyle f '(x)=\frac{3(x)}{1+3^2(x)}$

Is that correct?

5. no

$\displaystyle f(x) = tan^{-1} 3x$

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

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

6. Thanks for that