Derivative of Tangent

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

Let us suppose that the function 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 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 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

Differentiation 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}