Photo by Vickie Kelly, 2001

London Bridge, Lake Havasu City, Arizona

Derivatives of Trigonometric Functions

π

2π

0

2π

−

π−

Consider the function ( )siny θ=

We could make a graph of the slope: θ slope

1−

0

1

0

1−Now we connect the dots! The resulting curve is a cosine curve.

( )sin cosd x xdx

=

→

π

2π

0

2π

−

π−

We can do the same thing for ( )cosy θ= θ slope

0

1

0

1−

0The resulting curve is a sine curve that has been reflected about the x-axis.

( )cos sind x xdx

= −

→

We can find the derivative of tangent x by using the quotient rule.

tand xdx

sincos

d xdx x

( )2

cos cos sin sincos

x x x xx

⋅ − ⋅ −

2 2

2

cos sincosx xx

+

2

1cos x

2sec x

( ) 2tan secd x xdx

=

→

Derivatives of the remaining trig functions can be determined the same way.

sin cosd x xdx

=

cos sind x xdx

= −

2tan secd x xdx

=

2cot cscd x xdx

= −

sec sec tand x x xdx

= ⋅

csc csc cotd x x xdx

= − ⋅

π

Lets try this: Find the following derivatives y = xsin x

y = sin x1+ cos x

y ' = sin x + xcos x

y ' =cos x 1+ cos x( )− −sin x( )sin x

1+ cos x( )2

y ' = cos x + cos2 x + sin2 x

1+ cos x( )2

y ' = cos x +1

1+ cos x( )2

y ' = 11+ cos x( )

Lets try this: Find f " π4

!

"#

$

%& if f x( ) = sec x

f ' x( ) = sec x tan xf " x( ) = sec x tan x tan x + sec2 xsec x

f " x( ) = sec x tan2 x + sec3 x

f " x( ) = sec x tan2 x + sec2 x( )f " π

4!

"#

$

%&= sec

π4

!

"#

$

%& tan2

π4

!

"#

$

%&+ sec2

π4

!

"#

$

%&

!

"#

$

%&

f " π4

!

"#

$

%&=

221+ 2

2

!

"#

$

%&

2!

"

##

$

%

&& =

22+82 2

=62

= 3 2