Derivative of Secant

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

Let us suppose that the function 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 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 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 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

Differentiation 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}