# How do you differentiate arc cot(-4tan(1/(1-3x^2)) ) using the chain rule?

Feb 4, 2018

 d/dx "arccot"( -4tan(1/(1-3x^2)) ) = (24xsec^2(1/(1-3x^2)))/((1-3x^2)^(2) (1+16tan^2(1/(1-3x^2)))

#### Explanation:

Let use define:

$y = \text{arccot} \left(- 4 \tan \left(\frac{1}{1 - 3 {x}^{2}}\right)\right)$

 {: (ul("Function"), ul("Derivative"), ul("Notes")), (x^n, nx^(n-1), n " constant (Power rule)"), (tanx, sec^2x, ), ("arccot"^(-1)x, -1/(1+x^2), ), (f(g(x)), f'(g(x)) \ g'(x),"(Chain rule)" ) :}

We can apply the chain rule to get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{1 + {\left(- 4 \tan \left(\frac{1}{1 - 3 {x}^{2}}\right)\right)}^{2}} \setminus \frac{d}{\mathrm{dx}} \left(- 4 \tan \left(\frac{1}{1 - 3 {x}^{2}}\right)\right)$
$\setminus \setminus \setminus \setminus \setminus = \frac{4}{1 + 16 {\tan}^{2} \left(\frac{1}{1 - 3 {x}^{2}}\right)} \setminus \frac{d}{\mathrm{dx}} \left(\tan \left(\frac{1}{1 - 3 {x}^{2}}\right)\right)$

Then, we apply the chain rule a second time:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4}{1 + 16 {\tan}^{2} \left(\frac{1}{1 - 3 {x}^{2}}\right)} \setminus {\sec}^{2} \left(\frac{1}{1 - 3 {x}^{2}}\right) \setminus \frac{d}{\mathrm{dx}} \left(\frac{1}{1 - 3 {x}^{2}}\right)$

$\setminus \setminus \setminus \setminus \setminus = \frac{4 {\sec}^{2} \left(\frac{1}{1 - 3 {x}^{2}}\right)}{1 + 16 {\tan}^{2} \left(\frac{1}{1 - 3 {x}^{2}}\right)} \setminus \frac{d}{\mathrm{dx}} \left(\frac{1}{1 - 3 {x}^{2}}\right)$

$\setminus \setminus \setminus \setminus \setminus = \frac{4 {\sec}^{2} \left(\frac{1}{1 - 3 {x}^{2}}\right)}{1 + 16 {\tan}^{2} \left(\frac{1}{1 - 3 {x}^{2}}\right)} \setminus \frac{d}{\mathrm{dx}} \left({\left(1 - 3 {x}^{2}\right)}^{- 1}\right)$

Then, we apply the chain rule a third time:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 {\sec}^{2} \left(\frac{1}{1 - 3 {x}^{2}}\right)}{1 + 16 {\tan}^{2} \left(\frac{1}{1 - 3 {x}^{2}}\right)} \setminus \left(- 1\right) {\left(1 - 3 {x}^{2}\right)}^{- 2} \frac{d}{\mathrm{dx}} \left(1 - 3 {x}^{2}\right)$

$\setminus \setminus \setminus \setminus \setminus = \frac{4 {\sec}^{2} \left(\frac{1}{1 - 3 {x}^{2}}\right)}{1 + 16 {\tan}^{2} \left(\frac{1}{1 - 3 {x}^{2}}\right)} \setminus \frac{- 1}{1 - 3 {x}^{2}} ^ \left(2\right) \left(- 6 x\right)$

 \ \ \ \ \ = (24xsec^2(1/(1-3x^2)))/((1-3x^2)^(2) (1+16tan^2(1/(1-3x^2)))