first and second derivatives of the function

**jamm89** · May 31st 2009, 07:04 AM

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

**Amer** · May 31st 2009, 07:26 AM

$f(x)=tan^-1(u(x))$

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

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

**HallsofIvy** · May 31st 2009, 07:34 AM

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.

**jamm89** · May 31st 2009, 08:24 AM

$f(x)=tan^-1(3(x))$

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

Is that correct?

**Amer** · May 31st 2009, 08:27 AM

no

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

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

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

**jamm89** · May 31st 2009, 08:36 AM

Thanks for that

Thread: first and second derivatives of the function

LinkBack

Thread Tools

Display

first and second derivatives of the function

Similar Math Help Forum Discussions

question regarding derivatives of arc function

derivatives of the function.

first and second derivatives of a function

derivatives of log function application

Derivatives of log function!

Search Tags