# How do you differentiate f(x)=cot(1/sqrt(x-3))  using the chain rule?

Dec 24, 2015

Applying a "two-degrees" chain rule.

Just remembering the concept: $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dv}} \frac{\mathrm{dv}}{\mathrm{dx}}$

#### Explanation:

Here, we can name $u = {\left(v\right)}^{- \frac{1}{2}}$ and $v = x - 3$

Also, we must remember that the rule to derivate $\cot a$ is

$\cot a = - a ' {\csc}^{2} a$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - u ' {\csc}^{2} u \cdot \left(- \frac{1}{2 {v}^{\frac{3}{2}}}\right) \cdot \left(1\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \left(- \frac{1}{2 {v}^{\frac{3}{2}}}\right) {\csc}^{2} \left({v}^{- \frac{1}{2}}\right) \cdot \left(- \frac{1}{2 {\left(x - 3\right)}^{\frac{3}{2}}}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{\csc}^{2} {\left(x - 3\right)}^{- \frac{1}{2}}}{2 {\left(x - 3\right)}^{\frac{3}{2}}} \left(- \frac{1}{2 {\left(x - 3\right)}^{\frac{3}{2}}}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{\csc}^{2} \left(\frac{1}{x - 3} ^ \left(\frac{1}{2}\right)\right)}{4 {\left(x - 3\right)}^{3}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{\csc}^{2} \left(\frac{1}{\sqrt{x - 3}}\right)}{4 {\left(x - 3\right)}^{3}}$