Rules for Differentiation

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

Think of a “derivative” as the slope of the tangent lineAt a given point of a function.

d

dxf( ) =slope

Other notation: y =3x

′y =3

(The derivative line is the slope of the line).

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

0d

cdx

example: 3y

0y

The derivative of a constant is zero.

But which tangent line for this function?

But which tangent line for this function?

Consider a secant line:

d

dxf (x) =lim

h→ 0

f x+h( )− f (x)

h

Definition of the derivative

Using the definition to find the derivative-Substitute x+h into the formula:

(because h goes to 0!)

2 22

0limh

x h xdx

dx h

=lim

h→ 0

x2 +2xh+h2( )−x

2

h=

2x+h1

2x

Using the definition to find the derivative-Substitute f(x) into the formula:Find

d

dx(x3 −1lim

h→ 0

[ xh 3−1] −x3 −1

h

d

dx(x3 −1)

Using the definition to find the derivative-Substitute f(x) into the formula:Find

d

dx(x3 −1lim

h→ 0

[ xh 3−1] −x3 −1

h

d

dx(x3 −1)

=lim

h→ 0

x3 +3x2h+3xh2 +h3 −1( )−x3 +1

h

=limh→0

(3x2 + 3xh + h2 ) = 3x2

Do Now: find the derivative of the functionUsing the definition:

f (x) =1x

d

dxf (x) =lim

h→ 0

f x+h( )− f (x)

h

d

dx

1

x

⎛

⎝⎜⎞

⎠⎟=lim

h→ 0

1x+h

⎛

⎝⎜

⎞

⎠⎟−

1x

⎛

⎝⎜⎞

⎠⎟

h

=limh→ 0

x−(x+h)x(x+h)

h=

−hx(x+h)

h=

−hxh(x+h)

=limh→0

−1

x2 + xh=

−1

x2

If we find derivatives with the difference quotient:

2 22

0limh

x h xdx

dx h

2 2 2

0

2limh

x xh h x

h

2x

3 33

0limh

x h xdx

dx h

3 2 2 3 3

0

3 3limh

x x h xh h x

h

23x

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

(Pascal’s Triangle)

2

4dx

dx

4 3 2 2 3 4 4

0

4 6 4limh

x x h x h xh h x

h

34x

2 3

We observe a pattern: 2x 23x 34x 45x 56x …

1n ndx nx

dx

examples:

4f x x

34f x x′

8y x

78y x

power rule

We observe a pattern: 2x 23x 34x 45x 56x …

d ducu c

dx dx

examples:

1n ndcx cnx

dx

constant multiple rule:

5 4 47 7 5 35d

x x xdx

When we used the difference quotient, we observed that since the limit had no effect on a constant coefficient, that the constant could be factored to the outside.

(Each term is treated separately)

d ducu c

dx dx

constant multiple rule:

sum and difference rules:

d du dvu v

dx dx dx d du dv

u vdx dx dx

4 12y x x 34 12y x′

4 22 2y x x

34 4dy

x xdx

(Each term is treated separately)

d ducu c

dx dx

constant multiple rule:

sum and difference rules:

d du dvu v

dx dx dx d du dv

u vdx dx dx

y =x2 +12 y =−x3 −2x +1

Examples:

(Each term is treated separately)

d ducu c

dx dx

constant multiple rule:

sum and difference rules:

d du dvu v

dx dx dx d du dv

u vdx dx dx

y =x2 +12y'=2x

y =−x3 −2x +1dydx

=−3x2 −2

Examples:

Find the first derivative of the following:

f (x) =(x−7)(x+ 3)

Find the first derivative of the following:

Can you guess the second derivative?

f (x) =(x−7)(x+ 3) =x2 −4x−21ddx

=2x−4

Find the first derivative of the following:

the second derivative is:

f (x) =(x−7)(x+ 3) =x2 −4x−21ddx

=2x−4

d

d2x=2

Now: find the derivative of the functionUsing the power rule:

Use exponents first!

f (x) =1x

Now: find the derivative of the functionUsing the power rule:

Put back in fraction form!

f (x) =1x

=x−1

f '(x) =−1x−2 =−1x2

Applied to trickier questions: Put in exponential form first!

5 x

2

x3

Applied to trickier questions: apply the rule:

5 x =5x12

2x3

=2x−3

Don’t forget to reduce the exponent by 1!

5 x =5x12

2x3

=2x−3

f '(x) =52x

−12 =

52 x

=5 x2x

d

dx(2x−3) =−6x−4 =

−6x4

Example:Find 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.)

4 22 2y x x

4 22 2y x x

2y

4 22 2y x x

2y

1y

4 22 2y x x

4 22 2y x x

First derivative (slope) is zero at:

0, 1, 1x

34 4dy

x xdx

product rule:

d dv duuv u v

dx dx dx Notice that this is not just the

product of two derivatives.

This is sometimes memorized as: d uv u dv v du

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 15x 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 15x x

quotient rule:

2

du dvv ud u dx dx

dx v v

or 2

u v du u dvd

v 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

Higher Order Derivatives:

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.

We will learn later what these higher order derivatives are used for.