Derivatives of Trigonometric Functions
Posted by admin on Tuesday, 3 June 2014
Using algebra similar to how we found the Power Rule, the shortcuts for taking the derivatives of trigonometric functions turn out to be:
$$ {\operatorname{d}\!y \over \operatorname{d}\!x} [ \sin x ] = \cos x $$
$$ {\operatorname{d}\!y \over \operatorname{d}\!x} [ \cos x ] = -\sin x $$
$$ {\operatorname{d}\!y \over \operatorname{d}\!x} [ \tan x ] = {\sec^2} x $$
$$ {\operatorname{d}\!y \over \operatorname{d}\!x} [ \cot x ] = -{\csc^2} x $$
$$ {\operatorname{d}\!y \over \operatorname{d}\!x} [ \sec x ] = \sec x \tan x $$
$$ {\operatorname{d}\!y \over \operatorname{d}\!x} [ \csc x ] = -\csc x \cot x $$