## Calculus Tutorial

*Differentiating trigonometric function*

#### Intro

(1) d/dx(sinx)= cosx

(2) d/dx(cosx)= -sinx

(3) d/dx(tanx)= -( secx)2

(4) d/dx(cotx)= -(cosecx)2

(5) d/dx(secx)= secx.tanx

(6) d/dx(cosecx)= -cosecx.cotx

#### Sample Problem

#### Solution

Let y=tan(3π/2+x)

dy/dx=(sec(3π/2+x))2— differentiate

dy/dx( at x=π/4)=(sec(3π/2+π/4)2— substituting x=π/4

=(sec(7π/4))2

= ( sec(2π-π/4))2 —simplify

=(sec(π/4))2

=1/(cos(π/4))2

=1/(1/2)

=2

# About The Author

Mathematics Teacher |

I was a teacher in Srilanka from 1992 to 2006. I took tuition from 1990 to 1991 in India. I do general work in Canada but I help to do homework to my daughters. I finished my teacher\'s training college diploma in Srilanka. I completed my B.Sc degree course at Bharayiar university in India. |

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