Colorado National MonumentGreg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 2003

AP Calculus AB/BC

3.3 Differentiation Rules, p. 116

If the derivative of a function is its slope, then for a constant function, the derivative must be zero.

0dc

dx Example 1: 3y

0y

The derivative of a constant is zero.

Constant Rule, p. 116

2 2 .We saw that if , then y x y x

This is part of a pattern.

1n ndx nx

dx

Examples 2a & 2b:

4f x x

34f x x

8y x

78y x

Power Rule, p. 116

The Power Rule works for both positive and negative powers

d ducu c

dx dx Example 3

1n ndcx cnx

dx

Constant Multiple Rule, p. 117

57dx

dx

where c is a constant and u is a differentiable function of x.

5 17 5x 435x

(Each term is treated separately)

Sum and Difference Rules p. 117

d du dvu v

dx dx dx d du dv

u vdx dx dx

4 12y x x 34 12y x

where c is a constant and u is a differentiable function of x.

where u and v are differentiable functions of x.

Constant Multiple Rule, p. 117

d ducu c

dx dx

Example 4a Example 4b

34 4dy

x xdx

4 22 2y x x

Example 5Find the horizontal tangents of: 4 22 2y x x

34 4dy

x xdx

Horizontal tangents occur when slope = zero.34 4 0x x

3 0x x

2 1 0x x

1 1 0x x x

0, 1, 1x

Plugging the x values into the original equation, we get:

2, 1, 1y y y

(The function is even, so we only get two horizontal tangents.)

Example 5 (cont.)

4 22 2y x x

Example 5 (cont.)

4 22 2y x x

2y

Example 5 (cont.)

4 22 2y x x

2y

1y

Example 5 (cont.)

4 22 2y x x

Example 5 (cont.)

4 22 2y x x

First derivative (slope) is zero at:

0, 1, 1x

34 4dy

x xdx

Example 5 (cont.)

Product Rule, p. 119

d dv duuv u v

dx dx dx Notice that this is not just the

product of two derivatives.

This is sometimes memorized as: duv uv vu

dx

2 33 2 5d

x x xdx

5 3 32 5 6 15d

x x x xdx

5 32 11 15d

x x xdx

4 210 33 15 x x

2 3x 26 5x 32 5x x 2x

4 2 2 4 26 5 18 15 4 10x x x x x

4 210 33 15 x x

Example 6:

Quotient Rule, p. 120

2

du dvv ud u dx dx

dx v v

or 2

u vu uv

dv v

3

2

2 5

3

d x x

dx x

2 2 3

22

3 6 5 2 5 2

3

x x x x x

x

Example 7:

v u′ v′u

v2

Example 8

Find the tangent’s to Newton’s Serpentine at the origin and at the point (1, 2). 2

4

1

x

yx

2

22

1 4 4 2

1

x x xy

x

2 2

22

4 4 8

1

x xy

x

2

22

4 4

1

xy

x

Example 8 (cont.)

2

22

4 4

1

xy

x

Next, find the slopes at the given points.

For the point (0, 0), plug in 0 for x.y′ will be the slope.

2

22

4 0 4

0 1

y

So, m = 4 and the equation for the tangent is y = 4x since the y-intercept is 0.

For the point (1, 2), plug in 1 for x.

2

22

4 1 4

1 1

y

0y

So, m = 0 and the equation for the tangent is y = 2 since the m is 0.

Example 9: Find the first four derivatives of

3 25 2 y x x23 10 y x x

6 10 y x

6y(4) 0y

Higher Order Derivatives, p.122

dyy

dx is the first derivative of y with respect to x.

2

2

dy d dy d yy

dx dx dx dx

is the second derivative.

(y double prime)

dyy

dx

is the third derivative.

4 dy y

dx is the fourth derivative.