Rules for Differentiation Colorado National Monument Greg Kelly, Hanford High School, Richland, Washington Photo by Vickie Kelly, 2003
Jan 02, 2016
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.