# how to find derivative of tanx

## Question:

How to find derivative of{eq}\text{ tan} x {/eq}?

## Using the Quotient Rule for Derivatives:

In order to be able to determine the derivative of a function that is the quotient of two expressions, it is determined by the derivative of the numerator multiplied by the expression of the denominator minus the derivative of the denominator by the expression of the numerator, all divided by the expression of the square denominator: {eq}\,{\left[ {\frac{{u\left( x \right)}}{{v\left( x \right)}}} \right]^\prime } = \frac{{u{{\left( x \right)}^\prime } \cdot v\left( x \right) - u\left( x \right) \cdot {v^\prime }\left( x \right)}}{{{{\left( {v\left( x \right)} \right)}^2}}}{\text{.}} {/eq}

{eq}\eqalign{ & {\text{We're going to find the derivative for the function }}\,f\left( x \right) = \tan x.{\text{ }} \cr & {\text{Then}}{\text{, rewriting the expression in terms of }}\,\sin x{\text{ and }}\,\cos x,{\text{ using}} \cr & {\text{the trigonometric identity }}\,\tan x = \frac{{\sin x}}{{\cos x}}{\text{:}} \cr & \,\,\,f\left( x \right) = \tan x = \frac{{\sin x}}{{\cos x}} \cr & {\text{Then:}} \cr & \,\,\,{f^\prime }\left( x \right) = {\left( {\frac{{\sin x}}{{\cos x}}} \right)^\prime } \cr & {\text{Using the quotient rule }}\,{\left[ {\frac{{u\left( x \right)}}{{v\left( x \right)}}} \right]^\prime } = \frac{{u{{\left( x \right)}^\prime } \cdot v\left( x \right) - u\left( x \right) \cdot {v^\prime }\left( x \right)}}{{{{\left( {v\left( x \right)} \right)}^2}}}{\text{:}} \cr & \,\,\,{f^\prime }\left( x \right) = \frac{{{{\left( {\sin x} \right)}^\prime } \cdot \cos x - \sin x{{\left( {\cos x} \right)}^\prime }}}{{{{\left( {\cos x} \right)}^2}}} \cr & {\text{Appliying the common derivatives }}{\left( {\sin u} \right)^\prime } = \cos u \cdot {u^\prime }{\text{ and }}{\left( {\cos u} \right)^\prime } = - \sin u \cdot {u^\prime }{\text{:}} \cr & \,\,\,{f^\prime }\left( x \right) = \frac{{\cos \cdot \cos x - \sin x \cdot \left( { - \sin x} \right)}}{{{{\left( {\cos x} \right)}^2}}} \cr & {\text{Simplifying:}} \cr & \,\,\,{f^\prime }\left( x \right) = \frac{{{{\cos }^2}x + {{\sin }^2}x}}{{{{\cos }^2}x}} \cr & {\text{Using trigonometric identity }}\,{\cos ^2}x + {\sin ^2}x = 1{\text{ }} \cr & \,\,\,{f^\prime }\left( x \right) = \frac{1}{{{{\cos }^2}x}} \cr & {\text{Appliying trigonometric identity }}\frac{1}{{{{\cos }^2}x}} = {\sec ^2}x{\text{:}} \cr & \,\,\,{f^\prime }\left( x \right) = {\sec ^2}x \cr & {\text{Therefore}}{\text{, the derivative of the function }}\,\tan x\,{\text{ is: }}\boxed{{{\sec }^2}x} \cr} {/eq} 