# How do you differentiate y=arc cot(x/5)?

Jul 12, 2017

I got

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{5} \left(\frac{1}{1 + {\left(\frac{x}{5}\right)}^{2}}\right)$

I'll assume you don't know that $\frac{d}{\mathrm{dx}} \left[a r c \cot \left(u \left(x\right)\right)\right] = - \frac{1}{1 + {u}^{2}} \frac{\mathrm{du}}{\mathrm{dx}}$.

$\cot y = \frac{x}{5}$

Then, by implicit differentiation:

$- {\csc}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{5}$

Therefore:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{5 {\csc}^{2} y}$

And since we defined $y = a r c \cot \left(\frac{x}{5}\right)$, we can use the identity

${\csc}^{2} y = 1 + {\cot}^{2} y$

to get

$\textcolor{b l u e}{\frac{\mathrm{dy}}{\mathrm{dx}}} = - \frac{1}{5} \frac{1}{1 + {\cot}^{2} y}$

$= \textcolor{b l u e}{- \frac{1}{5} \left(\frac{1}{1 + {\left(\frac{x}{5}\right)}^{2}}\right)}$

If you did it using the actual derivative of $a r c \cos u$, you would still get

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{1 + {\underbrace{{\left(\frac{x}{5}\right)}^{2}}}_{{u}^{2}}} \cdot \frac{d}{\mathrm{dx}} \left[\frac{x}{5}\right]$

$= - \frac{1}{5} \left(\frac{1}{1 + {\left(\frac{x}{5}\right)}^{2}}\right)$