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.