# How do you differentiate f(x)=xsecx using the product rule?

$f ' \left(x\right) = \sec x + x \cdot \sec x \tan x$

#### Explanation:

We start from the given equation
$f \left(x\right) = x \sec x$

We will use the derivative of product formula

$\frac{d}{\mathrm{dx}} \left(u v\right) = v \cdot \frac{d}{\mathrm{dx}} \left(u\right) + u \cdot \frac{d}{\mathrm{dx}} \left(v\right)$

Let $u = x$ and $v = \sec x$

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \frac{d}{\mathrm{dx}} \left(x \sec x\right) = \sec x \cdot \frac{d}{\mathrm{dx}} \left(x\right) + x \cdot \frac{d}{\mathrm{dx}} \left(\sec x\right)$

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \left(\sec x\right) \left(1\right) + x \left(\sec x \tan x\right) \frac{d}{\mathrm{dx}} \left(x\right)$

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \sec x + x \left(\sec x \tan x\right) \cdot 1$

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \sec x + x \cdot \sec x \tan x$

God bless....I hope the explanation is useful