# Differentiate cot(xy)=k?

Dec 23, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{y}{x}$

#### Explanation:

This uses implicit differentiation, which is a special case of the chain rule for derivatives. Generally differentiation problems involve functions i.e. $y = f \left(x\right)$ - written explicitly as functions of $x$.

However, some functions of $y$ are written implicitly as functions of $x$ and either we cannot separate $y$ or doing so makes it complicated. Observe that $f \left(x , y\right) = \cot \left(x y\right) = k$, where$k$ is a constant is such a function.

So what we do is to treat $y$ as $y = y \left(x\right)$ and use chain rule. This means differentiating $y$ w.r.t. $y$, but as we have to derive w.r.t. $x$, as per chain rule, we multiply it by $\frac{\mathrm{dy}}{\mathrm{dx}}$.

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

${\csc}^{2} \left(x y\right) \left(1 \cdot y + x \cdot \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

or $y {\csc}^{2} \left(x y\right) + x {\csc}^{2} \left(x y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

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