# Question #fc13e

Oct 23, 2016

Let $y = a r \mathcal{o} t x$

Then $x = \cot y$.

Differentiate implicitly to get

$1 = - {\csc}^{2} y \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

So that $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\csc} ^ 2 y$

Recall from trigonometry that $1 + {\cot}^{2} y = {\csc}^{2} y$ so

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{1 + {\cot}^{2} y}$

Finally replace $\csc y$ with $x$ (from the second line of the explanation.)

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{1 + {x}^{2}}$

Oct 23, 2016

#### Explanation:

We'll use implicit differentiation:

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

$\implies \cot \left(y\right) = x$

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

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

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

Now let's see what ${\csc}^{2} \left(y\right)$ is in terms of $x$. Draw a right triangle with an angle $y$ such that $\cot \left(y\right) = x$ (i.e. $y = \text{arccot} \left(x\right)$):

Setting the legs so that $\cot \left(y\right) = x$, we can find the length of the hypotenuse as $\sqrt{1 + {x}^{2}}$ using the Pythagorean theorem. Now we can calculate ${\csc}^{2} \left(y\right)$.

${\csc}^{2} \left(y\right) = {\left(\csc \left(y\right)\right)}^{2} = {\left(\frac{\sqrt{1 + {x}^{2}}}{1}\right)}^{2} = 1 + {x}^{2}$

Substituting this in, we arrive at our answer:

$\left(\text{arccot} \left(x\right)\right) ' = \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{1 + {x}^{2}}$