Answer

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**Hint:**For this question, we will directly use the formula of first derivative principle to find the derivative. The formula is given below as: \[f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\]. So, use this concept to reach the solution of the given problem.

**Complete step by step answer:**

We have to find out the derivative of \[\tan x\]. So, our function will be \[f\left( x \right) = \tan x\].

According to the first derivative principle, we have

\[

\Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\tan \left( {x + h} \right) - \tan x}}{h} \\

\Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{{\sin \left( {x + h} \right)}}{{\cos \left( {x + h} \right)}} - \dfrac{{\sin x}}{{\cos x}}}}{h}{\text{ }}\left[ {\because \tan A = \dfrac{{\sin A}}{{\cos A}}} \right] \\

\]

Taking LCM and simplifying further, we have

\[ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin \left( {x + h} \right)\cos x - \cos \left( {x + h} \right)\sin x}}{{h\cos x\cos \left( {x + h} \right)}}\]

By using the formula, \[\sin A\cos B - \cos A\sin B = \sin \left( {A - B} \right)\], we have

\[

\Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin \left( {x + h - x} \right)}}{{h\cos x\cos \left( {x + h} \right)}} \\

\Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin \left( h \right)}}{{h\cos x\cos \left( {x + h} \right)}} \\

\]

Splitting the limits, we have

\[ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin \left( h \right)}}{h} \times \mathop {\lim }\limits_{h \to 0} \dfrac{1}{{\cos x\cos \left( {x + h} \right)}}\]

By using the formula, \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1\], we have

\[ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{1}{{\cos x\cos \left( {x + h} \right)}}\]

We know that for \[h \to 0\] we have \[\cos \left( {x + h} \right) \simeq \cos x\]

\[

\Rightarrow f'\left( x \right) = \dfrac{1}{{\cos x\cos x}} \\

\Rightarrow f'\left( x \right) = \dfrac{1}{{{{\cos }^2}x}} \\

\therefore f'\left( x \right) = {\sec ^2}x\,{\text{ }}\left[ {\because \sec x = \dfrac{1}{{\cos x}}} \right] \\

\]

**Thus, the derivative of \[\tan x\] using the first derivative principle is \[{\sec ^2}x\].**

**Note:**Using the first derivative method, it consumes much time. And for smaller functions, we can find out the derivative using the first derivative method. But if the function is complex, then it is too difficult to solve using this method. Then we follow conventional methods for finding the derivative.

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