# How do I find the derivative of y = arc cot(x) + (cot x)^-1?

Aug 29, 2015

${y}^{'} = {\sec}^{2} x - \frac{1}{1 + {x}^{2}}$

#### Explanation:

I'll assume that you don't know what the derivative of $\text{arccot} \left(x\right)$ is, so that we can use implicit differentiation to find $\frac{\mathrm{dy}}{\mathrm{dx}}$.

First, start by rewriting your function as

$y = \text{arccot"(x) + 1/cotx" }$, which is equivalent to

$y = \text{arccot} \left(x\right) + \tan \left(x\right)$

Rerrange to get $\arccos \left(x\right)$ alone on the right-hand side of the equation

$y - \tan \left(x\right) = \text{arccot} \left(x\right)$

This is equivalent to

$\cot \left(y - \tan \left(x\right)\right) = x \text{ } \textcolor{b l u e}{\left(1\right)}$

Differentiate both sides with respect to $x$ to get

$\frac{d}{\mathrm{dx}} \cot \left(y - \tan \left(x\right)\right) = \frac{d}{\mathrm{dx}} \left(x\right)$

$- {\csc}^{2} \left(y - \tan \left(x\right)\right) \cdot \left[\frac{\mathrm{dy}}{\mathrm{dx}} - {\sec}^{2} \left(x\right)\right] = 1$

$- {\csc}^{2} \left(y - \tan \left(x\right)\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + {\csc}^{2} \left(y - \tan \left(x\right)\right) \cdot {\sec}^{2} \left(x\right) = 1$

Rearrange to get $\frac{\mathrm{dy}}{\mathrm{dx}}$ alone on the left-hand side of the equation

$- {\csc}^{2} \left(y - \tan \left(x\right)\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 1 - {\csc}^{2} \left(y - \tan \left(x\right)\right) \cdot {\sec}^{2} \left(x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - {\csc}^{2} \left(y - \tan \left(x\right)\right) \cdot {\sec}^{2} \left(x\right)}{- {\csc}^{2} \left(y - \tan \left(x\right)\right)}$

(dy)/dx = -1/csc^2(y - tan(x)) + (color(red)(cancel(color(black)(-csc^2(y - tan(x))))) * sec^2(x))/color(red)(cancel(color(black)(-csc^2(y - tan(x))))

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\sec}^{2} x - \frac{1}{\csc} ^ 2 \left(y - \tan \left(x\right)\right)$

Use the trigonometric identity

$\textcolor{b l u e}{{\csc}^{2} x = 1 + {\cot}^{2} x}$

to write

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\sec}^{2} x - \frac{1}{1 + {\cot}^{2} \left(y - \tan \left(x\right)\right)}$

You know from equation $\textcolor{b l u e}{\left(1\right)}$ that

$\cot \left(y - \tan \left(x\right)\right) = x$

which means that the derivative will be

$\frac{\mathrm{dy}}{\mathrm{dx}} = \textcolor{g r e e n}{{\sec}^{2} x - \frac{1}{1 + {x}^{2}}}$