To find the derivative of $\mathrm{sec}(x)$, we start by rewriting it in a form that might be easier to differentiate. Recall that:

$$\mathrm{sec}(x)={\displaystyle \frac{1}{\mathrm{cos}(x)}}$$To differentiate this, we use the <strong>quotient rule</strong>. The quotient rule states that for two functions $u(x)$ and $v(x)$, the derivative of their quotient $\frac{u}{v}$ is given by:

$$\frac{\text{d}}{\text{d}x}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{{v}^{2}}$$In our case:

- $u=1$ (the numerator)
- $v=\mathrm{cos}(x)$ (the denominator)

**Step 1: Differentiate $u$ and $v$**

- The derivative of $u$ with respect to $x$ is $0$ because $u=1$ – a constant.
- The derivative of $v=\mathrm{cos}(x)$ with respect to $x$ is $-\mathrm{sin}(x)$.

**Step 2: Apply the quotient rule**

Plugging these into the quotient rule:

$$\frac{\text{d}}{\text{d}x}\left(\frac{1}{\mathrm{cos}(x)}\right)=\frac{\mathrm{cos}(x)\xb70-1\xb7(-\mathrm{sin}(x))}{{\mathrm{cos}}^{2}(x)}$$This simplifies to:

$$\frac{\mathrm{sin}(x)}{{\mathrm{cos}}^{2}(x)}$$**Step 3: Simplify the result**

Now, we can rewrite this expression in terms of other trigonometric functions:

$$\frac{\mathrm{sin}(x)}{{\mathrm{cos}}^{2}(x)}=\frac{\mathrm{sin}(x)}{\mathrm{cos}(x)}\xb7\frac{1}{\mathrm{cos}(x)}=\mathrm{tan}(x)\xb7\mathrm{sec}(x)$$Thus, the derivative of $\mathrm{sec}(x)$ is:

$$\overline{)\mathrm{sec}(x)\xb7\mathrm{tan}(x)}$$*Q.E.D.*