Derivative of Secant Squared Function

In this tutorial we shall discuss derivative of secant squared function and its related examples. It can be proved by definition of differentiation.

We have function of the form

y = f\left( x \right) = {\sec ^2}x

By the definition of differentiation we have

\frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{f\left( {x + \Delta x} \right) - f\left( x \right)}}{{\Delta x}}\,\,\,\,{\text{ - - - }}\left( {\text{i}} \right)

Putting the value of function in equation (i), we get

\begin{gathered} \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{{{\sec }^2}\left( {x + \Delta x} \right) - {{\sec }^2}x}}{{\Delta x}} \\ \Rightarrow \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\left[ {\sec \left( {x + \Delta x} \right) + \sec x} \right]\left[ {\sec \left( {x + \Delta x} \right) - \sec x} \right]}}{{\Delta x}} \\ \Rightarrow \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\left[ {\sec \left( {x + \Delta x} \right) + \sec x} \right]}}{{\Delta x}}\left[ {\frac{1}{{\cos \left( {x + \Delta x} \right)}} - \frac{1}{{\cos x}}} \right] \\ \Rightarrow \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\left[ {\sec \left( {x + \Delta x} \right) + \sec x} \right]}}{{\Delta x}}\left[ {\frac{{\cos x - \cos \left( {x + \Delta x} \right)}}{{\cos \left( {x + \Delta x} \right)\cos x}}} \right] \\ \Rightarrow \frac{{dy}}{{dx}} = - \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\left[ {\sec \left( {x + \Delta x} \right) + \sec x} \right]}}{{\cos \left( {x + \Delta x} \right)\cos x\left( {\Delta x} \right)}}\left[ {\cos \left( {x + \Delta x} \right) - \cos x} \right] \\ \end{gathered}

Using formula from trigonometry

\cos A - \cos B = - 2\sin \left( {\frac{{A + B}}{2}} \right)\sin \left( {\frac{{A - B}}{2}} \right)

\begin{gathered} \frac{{dy}}{{dx}} = - \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\left[ {\sec \left( {x + \Delta x} \right) + \sec x} \right]}}{{\cos \left( {x + \Delta x} \right)\cos x\left( {\Delta x} \right)}}\left[ { - 2\sin \left( {\frac{{x + \Delta x + x}}{2}} \right)\sin \left( {\frac{{x + \Delta x - x}}{2}} \right)} \right] \\ \Rightarrow \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\left[ {\sec \left( {x + \Delta x} \right) + \sec x} \right]}}{{\cos \left( {x + \Delta x} \right)\cos x}}\sin \left( {\frac{{2x + \Delta x}}{2}} \right)\frac{{\sin \left( {\frac{{\Delta x}}{2}} \right)}}{{\frac{{\Delta x}}{2}}} \\ \Rightarrow \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\left[ {\sec \left( {x + \Delta x} \right) + \sec x} \right]}}{{\cos \left( {x + \Delta x} \right)\cos x}}\mathop {\lim }\limits_{\Delta x \to 0} \sin \left( {\frac{{2x + \Delta x}}{2}} \right)\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin \left( {\frac{{\Delta x}}{2}} \right)}}{{\frac{{\Delta x}}{2}}} \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{{\left[ {\sec \left( {x + 0} \right) + \sec x} \right]}}{{\cos \left( {x + 0} \right)\cos x}}\sin \left( {\frac{{2x + 0}}{2}} \right)\left( 1 \right) \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{{2\sec x}}{{{{\cos }^2}x}}\sin x \\ \Rightarrow \frac{{dy}}{{dx}} = 2\sec x\sec x\tan x \\ \Rightarrow \frac{{dy}}{{dx}} = 2{\sec ^2}x\tan x \\ \end{gathered}

Example: Find the derivative of

y = f\left( x \right) = {\sec ^2}3x

We have the given function as

y = {\sec ^2}3x

Differentiation with respect to variable x, we get

\frac{{dy}}{{dx}} = \frac{d}{{dx}}{\sec ^2}3x

Using the rule, \frac{d}{{dx}}{\sec ^2}x = 2{\sec ^2}x\tan x, we get

\begin{gathered} \frac{{dy}}{{dx}} = 2{\sec ^2}3x\tan 3x\frac{d}{{dx}}3x \\ \Rightarrow \frac{{dy}}{{dx}} = 6{\sec ^2}3x\tan 3x \\ \end{gathered}