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Mullins Beck

12/08/2020 · Junior High School

1) Use the limit definition of the derivative \(( \lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } ) \) together with the tangent addition formula to prove that \(\frac { d } { d x } [ \tan x ] = \sec ^ { 2 } x\) 

Tangent Addition Formula: \(\tan ( x + y ) = \frac { \tan x + \tan y } { 1 - \tan x \tan y } \) 

You may also use this graph: \(f ( x ) = \frac { \tan x } { x } \) 

Answer
expertExpert-Verified Answer

Alexander Patel
Certificated Tutor
4.0 (47votes)

Prelimilaries: \(\lim _ { h \to 0} \frac { \tan h} { h} = 0\)\(\lim _ { h \to 0} 1 - \tan x \tan h = 1\).

By using the prelimilaries, we have:

 

\(( \tan x) ' = \lim _ { h \to 0} \frac { \tan ( x+ h) - \tan ( x) } { h} = \lim _ { h \to 0} \frac { \frac { \tan ( x) + \tan ( h) } { 1 - \tan ( x) \tan ( h) } - \tan ( x) } { h} = \lim _ { h \to 0} \frac { \tan ( x) + \tan ( h) - \tan ( x) + \tan ^ 2( x) \tan ( h) } { h( 1- \tan ( x) \tan ( h) ) } = \lim _ { h \to 0} \frac { 1 + \tan ^ 2( x) } { 1 - \tan ( x) \tan ( h) } \cdot \frac { \tan ( h) } { h} = 1 + \tan ^ 2( x) = \sec ^ 2x\)

 

 

 

 

 

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