# How do you differentiate g(x) = sqrt(x^2-1)cot(7-x) using the product rule?

Jun 22, 2016

$\frac{d}{\mathrm{dx}} \left(\sqrt{{x}^{2} - 1} \cot \left(7 - x\right)\right) = \frac{x \cot \left(- x + 7\right)}{\setminus \sqrt{{x}^{2} - 1}} + \frac{\sqrt{{x}^{2} - 1}}{{\sin}^{2} \left(- x + 7\right)}$

#### Explanation:

$\frac{d}{\mathrm{dx}} \left(\sqrt{{x}^{2} - 1} \cot \left(7 - x\right)\right)$

Applying product rule, ${\left(f \cdot g\right)}^{'} = {f}^{'} \cdot g + f \cdot {g}^{'}$

$f = \sqrt{{x}^{2} - 1} , g = \cot \left(7 - x\right)$

$= \frac{d}{\mathrm{dx}} \left(\sqrt{{x}^{2} - 1}\right) \cot \left(7 - x\right) + \frac{d}{\mathrm{dx}} \left(\cot \left(7 - x\right)\right) \sqrt{{x}^{2} - 1}$

We know,
$\frac{d}{\mathrm{dx}} \left(\sqrt{{x}^{2} - 1}\right) = \frac{x}{\sqrt{{x}^{2} - 1}}$

$\frac{d}{\mathrm{dx}} \left(\cot \left(7 - x\right)\right) = \frac{1}{{\sin}^{2} \left(7 - x\right)}$

[Applying chain rule $\frac{\mathrm{df} \left(u\right)}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

let $7 - x = u$

$= \frac{d}{\mathrm{du}} \left(\cot \left(u\right)\right) \frac{d}{\mathrm{dx}} \left(7 - x\right)$

we know,$\frac{d}{\mathrm{du}} \left(\cot \left(u\right)\right) = - \frac{1}{{\sin}^{2} \left(u\right)}$
and,$\frac{d}{\mathrm{dx}} \left(7 - x\right) = - 1$

so,$= \frac{1}{{\sin}^{2} \left(7 - x\right)}$]

Finally,
$= \frac{x}{\sqrt{{x}^{2} - 1}} \cot \left(7 - x\right) + \frac{1}{{\sin}^{2} \left(7 - x\right)} \sqrt{{x}^{2} - 1}$

simplifying it,
$= \frac{x \cot \left(- x + 7\right)}{\setminus \sqrt{{x}^{2} - 1}} + \frac{\sqrt{{x}^{2} - 1}}{{\sin}^{2} \left(- x + 7\right)}$