Question

prove for all trigonometric differentiation

Answer 1

FunctionDerivative{\displaystyle \sin(x)}{\displaystyle \cos(x)}{\displaystyle \cos(x)}{\displaystyle -\sin(x)}{\displaystyle \tan(x)}{\displaystyle \sec ^{2}(x)}{\displaystyle \cot(x)}{\displaystyle -\csc ^{2}(x)}{\displaystyle \sec(x)}{\displaystyle \sec(x)\tan(x)}{\displaystyle \csc(x)}{\displaystyle -\csc(x)\cot(x)}{\displaystyle \arcsin(x)}{\displaystyle {\frac {1}{\sqrt {1-x^{2}}}}}{\displaystyle \arccos(x)}{\displaystyle -{\frac {1}{\sqrt {1-x^{2}}}}}{\displaystyle \arctan(x)}{\displaystyle {\frac {1}{x^{2}+1}}}{\displaystyle \operatorname {arccot} (x)}{\displaystyle -{\frac {1}{x^{2}+1}}}{\displaystyle \operatorname {arcsec}(x)}{\displaystyle {\frac {1}{|x|{\sqrt {x^{2}-1}}}}}{\displaystyle \operatorname {arccsc}(x)}{\displaystyle -{\frac {1}{|x|{\sqrt {x^{2}-1}}}}}

The differentiation of trigonometric functionsis the mathematical process of finding thederivative of a trigonometric function, or its rate of change with respect to a variable. Common trigonometric functions include sin(x), cos(x) and tan(x). For example, the derivative of f(x) = sin(x) is represented as f ′(a) = cos(a). f ′(a) is the rate of change of sin(x) at a particular point a.

All derivatives of circular trigonometric functions can be found using those of sin(x) and cos(x). The quotient rule is then implemented to differentiate the resulting expression. Finding the derivatives of theinverse trigonometric functions involves usingimplicit differentiation and the derivatives of regular trigonometric functions.