# How do you differentiate g(x) = sqrt(x^2-3)sec(2-x) using the product rule?

Apr 12, 2018

$g ' \left(x\right) = \sec \left(2 - x\right) \frac{x - \left({x}^{2} - 3\right) \tan \left(2 - x\right)}{\sqrt{{x}^{2} - 3}}$

#### Explanation:

The product rule says

$\frac{\mathrm{df} \left(x\right) h \left(x\right)}{\mathrm{dx}} = f ' \left(x\right) h \left(x\right) + f \left(x\right) h ' \left(x\right)$

Here

$f \left(x\right) = \sqrt{{x}^{2} - 3}$

$f ' \left(x\right) = \frac{2 x}{2 \sqrt{{x}^{2} - 3}} = \frac{x}{\sqrt{{x}^{2} - 3}}$ (by the chain rule), and

$h \left(x\right) = \sec \left(2 - x\right)$

$h ' \left(x\right) = - \tan \left(2 - x\right) \sec \left(2 - x\right)$ (by the chain rule).

Therefore,

$g ' \left(x\right) = \frac{x \sec \left(2 - x\right)}{\sqrt{{x}^{2} - 3}} - \sqrt{{x}^{2} - 3} \tan \left(2 - x\right) \sec \left(2 - x\right)$

$= \sec \left(2 - x\right) \left(\frac{x}{\sqrt{{x}^{2} - 3}} - \sqrt{{x}^{2} - 3} \tan \left(2 - x\right)\right)$

$= \sec \left(2 - x\right) \frac{x - \left({x}^{2} - 3\right) \tan \left(2 - x\right)}{\sqrt{{x}^{2} - 3}}$