VCE Maths Methods - Derivatives of circular functions

Differentiating sine functions

1

• The derivative of the sine function is the cosine function.

gradient = 0

gradient = 0

gradient = -1

cosπ =−1

cos 3π

2=0

cosπ

2=0

cos0=1

gradient = 1

y = sin (x)y = cos (x)

VCE Maths Methods - Derivatives of circular functions

Differentiating sine & cosine functions

2

• The derivative of the sin function is the cosine function.

ddx

sin(x )= cos(x )

• The derivative of the cosine function is the (negative) sine function.

ddx

cos(x )=−sin(x )

VCE Maths Methods - Derivatives of circular functions

Differentiating sine functions with different periods

3

• If the sin function has a dilation parallel to the x axis by the factor k, then there will be k periods per 2π and the gradient will be k times greater at the corresponding point along the curve.

y =sin(x ) y =sin(2x )

ddx

sin(2x )=2cos(2x )

VCE Maths Methods - Derivatives of circular functions

Differentiating sine functions with different periods

4

• Use the chain rule to differentiate:

y =sin(kx )y =sin(u),u =kx

ddx

sin(kx )=kcos(kx )

dydx

= dydu

× dudx

= cox(kx )×k

y =sin(kx )

dydu

= cos(u),dudx

=k

ddx

cos(kx )=−ksin(kx )

dydx

= dydu

⋅ dudx

VCE Maths Methods - Derivatives of circular functions

Differentiating tangent functions

5

• The tan function is found from the ratio of sine and cosine functions.

ddx

tan(x )= ddx

sin(x )cos(x )

= 1cos2(x )

=sec2(x ) (Using the quotient rule)

ddx

tanx = 1cos2x

=sec2x

ddx

tankx = kcos2x

=ksec2x (Using the chain rule)

VCE Maths Methods - Derivatives of circular functions

Differentiating tangent functions

6

y =tan(x )

y = 1

cos2(x )

VCE Maths Methods - Derivatives of circular functions

Differentiating circular functions (1)

7

• Find the derivative of the function: f (x )=4sin(2πx )

f '(x )= (4×2π )cos(2πx )

f '(x )=8πcos(2πx )

ddx

sin(kx )=kcos(kx )

VCE Maths Methods - Derivatives of circular functions

Differentiating circular functions (2)

8

y =4cos2(x +π

6)−5• Find the derivative of the function:

y =4cos(u)−5 u =2(x +π

6)=2x +π

3

dydu

=−4sin(u)

dudx

=2

dydx

= dydu

× dudx

=−4sin(u)×2

dydx

=−8sin2(x +π6

)

dydx

= dydu

⋅ dudx

VCE Maths Methods - Derivatives of circular functions

Differentiating circular functions (3)

9

f (x )=−3tan

x2

⎛⎝

⎞⎠ +1• Find the derivative of the function:

f (x )=−3tan

x2

⎛⎝

⎞⎠ +1

f '(x )=− 3

2cos2 x2

⎛⎝

⎞⎠

f '(x )=−12× 3

cos2 x2

⎛⎝

⎞⎠

ddx

tankx = kcos2x

VCE Maths Methods - Derivatives of circular functions

Differentiating circular functions (4)

10

y =3sin(x 2 −3x )• Find the derivative of the function:

y =3sin(u) u = x 2 −3x

dydu

=3cos(u)

dudx

=2x −3

dydx

= dydu

× dudx

=3cos(u)×(2x −3)

dydx

=3(2x −3)cos(x 2 −3x )

dydx

= dydu

⋅ dudx