# How do you differentiate f(x)=tan^-2(1/(3x-1)) using the chain rule?

Apr 15, 2017

$\left(- 2 {\left(\tan \left(\frac{1}{3 x - 1}\right)\right)}^{-} 3\right) \left({\sec}^{2} \left(\left(\frac{1}{3} x - 1\right)\right)\right) \left(- {\left(3 x - 1\right)}^{-} 2\right) \left(3\right)$

#### Explanation:

There are $4$ "chains" in this case.

${u}^{-} 2$ where $u = \tan \left(\frac{1}{3 x - 1}\right)$

$\tan \left(w\right)$ where $w = \left(\frac{1}{3} x - 1\right)$

${v}^{-} 1$ where $v = 3 x - 1$

3x-1

We need to take the derivatives of each of these "chains"

$\left({u}^{-} 2\right) ' = - 2 {u}^{-} 3$

$\left(\tan \left(w\right)\right) ' = {\sec}^{2} \left(w\right)$

$\left(\frac{1}{v}\right) ' = - {v}^{-} 2$

$\left(3 x - 1\right) ' = 3$

Now multiply all these components together:

$\left(- 2 {u}^{-} 3\right) \left({\sec}^{2} \left(w\right)\right) \left(- {v}^{-} 2\right) \left(3\right)$

Now substitute back in

(-2(tan(1/(3x-1)))^-3)(sec^2((1/(3x-1)))(-(3x-1)^-2)(3)

We can clean this up by moving the tangent and the ${\left(- 3 x - 1\right)}^{-} 2$ to the denominator and making the exponent positive and then multiplying all the coefficients:

(6sec^2(1/(3x-1)))/((3x-1)^2(tan^3(1/(3x-1)))#