# Derivative of Secant

We shall prove the formula for the derivative of the secant function using definition or the first principle method.

Let us suppose that the function is of the form $y = f\left( x \right) = \sec x$

First we take the increment or small change in the function:
$\begin{gathered} y + \Delta y = \sec \left( {x + \Delta x} \right) \\ \Delta y = \sec \left( {x + \Delta x} \right) – y \\ \end{gathered}$

Putting the value of function $y = \sec x$ in the above equation, we get
$\begin{gathered} \Delta y = \sec \left( {x + \Delta x} \right) – \sec x \\ \Delta y = \frac{1}{{\cos \left( {x + \Delta x} \right)}} – \frac{1}{{\cos x}} \\ \end{gathered}$

Taking the LCM (Least Common Factor), we get
$\begin{gathered} \Delta y = \frac{{\cos x – \cos \left( {x + \Delta x} \right)}}{{\cos \left( {x + \Delta x} \right)\cos x}} \\ \Delta y = – \frac{{\cos \left( {x + \Delta x} \right) – \cos x}}{{\cos \left( {x + \Delta x} \right)\cos x}}\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right) \\ \end{gathered}$

Using the formula from trigonometry, we have
$\cos A – \cos B = – 2\sin \left( {\frac{{A + B}}{2}} \right)\sin \left( {\frac{{A – B}}{2}} \right)$

Using this formula in equation (i), we get
$\begin{gathered}\Delta y = – \frac{{\left[ { – 2\sin \left\{ {\frac{{\left( {x + \Delta x} \right) + x}}{2}} \right\}\sin \left\{ {\frac{{\left( {x + \Delta x} \right) – x}}{2}} \right\}} \right]}}{{\cos \left( {x + \Delta x} \right)\cos x}} \\ \Delta y = \frac{{2\sin \left\{ {\frac{{2x + \Delta x}}{2}} \right\}\sin \left\{ {\frac{{\Delta x}}{2}} \right\}}}{{\cos \left( {x + \Delta x} \right)\cos x}} \\ \end{gathered}$

Dividing both sides by $\Delta x$, we get
$\begin{gathered}\Rightarrow \frac{{\Delta y}}{{\Delta x}} = \frac{{2\sin \left\{ {\frac{{2x + \Delta x}}{2}} \right\}\sin \left\{ {\frac{{\Delta x}}{2}} \right\}}}{{\Delta x\cos \left( {x + \Delta x} \right)\cos x}} \\ \Rightarrow \frac{{\Delta y}}{{\Delta x}} = \frac{{\sin \left\{ {\frac{{2x + \Delta x}}{2}} \right\}}}{{\cos \left( {x + \Delta x} \right)\cos x}} \times \frac{{\sin \left\{ {\frac{{\Delta x}}{2}} \right\}}}{{\frac{{\Delta x}}{2}}} \\ \end{gathered}$

Taking the limit of both sides as $\Delta x \to 0$, we have
$\begin{gathered} \Rightarrow \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin \left\{ {\frac{{2x + \Delta x}}{2}} \right\}}}{{\cos \left( {x + \Delta x} \right)\cos x}} \times \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin \left\{ {\frac{{\Delta x}}{2}} \right\}}}{{\frac{{\Delta x}}{2}}} \\ \Rightarrow \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \frac{{\sin \left\{ {\frac{{2x + 0}}{2}} \right\}}}{{\cos \left( {x + 0} \right)\cos x}} \times \left( 1 \right) \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{{\sin x}}{{{{\cos }^2}x}} \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{\cos x}} \cdot \frac{{\sin x}}{{\cos x}} \\ \Rightarrow \frac{{dy}}{{dx}} = \sec x\tan x \\ \Rightarrow \frac{d}{{dx}}\sec x = \sec x\tan x \\ \end{gathered}$

Example: Find the derivative of $y = f\left( x \right) = \sec 5x$

We have the given function as
$y = \sec 5x$

Differentiating with respect to variable $x$, we get
$\frac{{dy}}{{dx}} = \frac{d}{{dx}}\sec 5x$

Using the rule, $\frac{d}{{dx}}\sec x = \sec x\tan x$, we get
$\begin{gathered}\frac{{dy}}{{dx}} = \sec 5x\tan 5x\frac{d}{{dx}}5x \\ \Rightarrow \frac{{dy}}{{dx}} = \sec 5x\tan 5x\left( 5 \right) \\ \Rightarrow \frac{{dy}}{{dx}} = 5\sec 5x\tan 5x \\ \end{gathered}$