Derivative of Tangent

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

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

First we take the increment or small change in the function:

\begin{gathered} y + \Delta y = \tan \left( {x + \Delta x} \right) \\ \Delta y = \tan \left( {x + \Delta x} \right) - y \\ \end{gathered}

Putting the value of function y = \tan x in the above equation, we get

\begin{gathered} \Delta y = \tan \left( {x + \Delta x} \right) - \tan x \\ \Delta y = \frac{{\sin \left( {x + \Delta x} \right)}}{{\cos \left( {x + \Delta x} \right)}} - \frac{{\sin x}}{{\cos x}} \\ \Delta y = \frac{{\sin \left( {x + \Delta x} \right)\cos x - \cos \left( {x + \Delta x} \right)\sin x}}{{\cos \left( {x + \Delta x} \right)\cos x}}\,\,\,\,{\text{ - - - }}\left( {\text{i}} \right) \\ \end{gathered}

Using the formula from trigonometry, we have

\sin \left( {\alpha - \beta } \right) = \sin \alpha \cos \beta - \cos \alpha \sin \beta

Using this formula in equation (i), we get

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

Example: Find the derivative of

y = f\left( x \right) = \tan 2x

We have the given function as

y = \tan 2x

Differentiating with respect to variable x, we get

\frac{{dy}}{{dx}} = \frac{d}{{dx}}\tan 2x

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

\begin{gathered}\frac{{dy}}{{dx}} = {\sec ^2}2x\frac{d}{{dx}}2x \\ \frac{{dy}}{{dx}} = {\sec ^2}2x \times 2\left( 1 \right) \\ \frac{{dy}}{{dx}} = 2{\sec ^2}2x \\ \end{gathered}