Differentiation Formulas Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example General Power Fun
Differentiation Formulas
As we did with limits and continuity, we will introduce several properties of the
derivative and use them along with the derivatives of some basic functions to
make calculation of derivatives easier.
Constant Functions and Power Functions
d
(c) = 0, if c is a constant.
Derivative of a Constatnt: dx
d
Power Rule : If n is a positive integer, then dx
(x n ) = nx n−1 .
3
0
Example If g (x) = 2, f (x) = x , find f (x) and g 0 (x).
Just as with limits, we have the following rules:
d
d
Constant Multiple Rule : dx
[cf (x)] = c dx
f (x), where c is a constant and
f is a differentiable function.
Example Let f (x) = x 3 , find f 0 (x), f 00 (x), f (3) (x) and f (4) (x).
Annette Pilkington
Lecture 23 : Sequences
Differentiation Formulas Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example General Power Fun
Differentiation Formulas
As we did with limits and continuity, we will introduce several properties of the
derivative and use them along with the derivatives of some basic functions to
make calculation of derivatives easier.
Constant Functions and Power Functions
d
(c) = 0, if c is a constant.
Derivative of a Constatnt: dx
d
Power Rule : If n is a positive integer, then dx
(x n ) = nx n−1 .
3
0
Example If g (x) = 2, f (x) = x , find f (x) and g 0 (x).
I
g 0 (x) = 0
Just as with limits, we have the following rules:
d
d
Constant Multiple Rule : dx
[cf (x)] = c dx
f (x), where c is a constant and
f is a differentiable function.
Example Let f (x) = x 3 , find f 0 (x), f 00 (x), f (3) (x) and f (4) (x).
Annette Pilkington
Lecture 23 : Sequences
Differentiation Formulas Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example General Power Fun
Differentiation Formulas
As we did with limits and continuity, we will introduce several properties of the
derivative and use them along with the derivatives of some basic functions to
make calculation of derivatives easier.
Constant Functions and Power Functions
d
(c) = 0, if c is a constant.
Derivative of a Constatnt: dx
d
Power Rule : If n is a positive integer, then dx
(x n ) = nx n−1 .
3
0
Example If g (x) = 2, f (x) = x , find f (x) and g 0 (x).
I
g 0 (x) = 0
I
f 0 (x) = 3x 2
Just as with limits, we have the following rules:
d
d
Constant Multiple Rule : dx
[cf (x)] = c dx
f (x), where c is a constant and
f is a differentiable function.
Example Let f (x) = x 3 , find f 0 (x), f 00 (x), f (3) (x) and f (4) (x).
Annette Pilkington
Lecture 23 : Sequences
Differentiation Formulas Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example General Power Fun
Differentiation Formulas
As we did with limits and continuity, we will introduce several properties of the
derivative and use them along with the derivatives of some basic functions to
make calculation of derivatives easier.
Constant Functions and Power Functions
d
(c) = 0, if c is a constant.
Derivative of a Constatnt: dx
d
Power Rule : If n is a positive integer, then dx
(x n ) = nx n−1 .
3
0
Example If g (x) = 2, f (x) = x , find f (x) and g 0 (x).
I
g 0 (x) = 0
I
f 0 (x) = 3x 2
Just as with limits, we have the following rules:
d
d
Constant Multiple Rule : dx
[cf (x)] = c dx
f (x), where c is a constant and
f is a differentiable function.
Example Let f (x) = x 3 , find f 0 (x), f 00 (x), f (3) (x) and f (4) (x).
I
f 0 (x) = 3x 2
Annette Pilkington
Lecture 23 : Sequences
Differentiation Formulas Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example General Power Fun
Differentiation Formulas
As we did with limits and continuity, we will introduce several properties of the
derivative and use them along with the derivatives of some basic functions to
make calculation of derivatives easier.
Constant Functions and Power Functions
d
(c) = 0, if c is a constant.
Derivative of a Constatnt: dx
d
Power Rule : If n is a positive integer, then dx
(x n ) = nx n−1 .
3
0
Example If g (x) = 2, f (x) = x , find f (x) and g 0 (x).
I
g 0 (x) = 0
I
f 0 (x) = 3x 2
Just as with limits, we have the following rules:
d
d
Constant Multiple Rule : dx
[cf (x)] = c dx
f (x), where c is a constant and
f is a differentiable function.
Example Let f (x) = x 3 , find f 0 (x), f 00 (x), f (3) (x) and f (4) (x).
I
f 0 (x) = 3x 2
I
f 00 (x) =
df 0 (x)
dx
=
d3x 2
dx
= 3(2x) = 6x.
Annette Pilkington
Lecture 23 : Sequences
Differentiation Formulas Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example General Power Fun
Differentiation Formulas
As we did with limits and continuity, we will introduce several properties of the
derivative and use them along with the derivatives of some basic functions to
make calculation of derivatives easier.
Constant Functions and Power Functions
d
(c) = 0, if c is a constant.
Derivative of a Constatnt: dx
d
Power Rule : If n is a positive integer, then dx
(x n ) = nx n−1 .
3
0
Example If g (x) = 2, f (x) = x , find f (x) and g 0 (x).
I
g 0 (x) = 0
I
f 0 (x) = 3x 2
Just as with limits, we have the following rules:
d
d
Constant Multiple Rule : dx
[cf (x)] = c dx
f (x), where c is a constant and
f is a differentiable function.
Example Let f (x) = x 3 , find f 0 (x), f 00 (x), f (3) (x) and f (4) (x).
I
f 0 (x) = 3x 2
I
f 00 (x) =
I
f (3) (x) =
df 0 (x)
dx
d6x
dx
=
d3x 2
dx
= 3(2x) = 6x.
= 6x 0 = 6.
f (4) (x) =
Annette Pilkington
d6
dx
= 0.
Lecture 23 : Sequences
Differentiation Formulas
Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example General Power Fun
Sums and Differences
The Sum Rule if f and g are both differentiable at x, then f + g is
differentiable at x and
d
d
d
[f (x) + g (x)] =
f (x) +
g (x)
dx
dx
dx
The Difference Rule if f and g are both differentiable at x, then f − g is
differentiable at x and
d
d
d
[f (x) − g (x)] =
f (x) −
g (x)
dx
dx
dx
Example
Find the derivative of the function f (x) = x 2 + 2x + 4.
Example
Find the derivative of the function f1 (x) = x 12 − 10x 6 + 3x + 1.
Annette Pilkington
Lecture 23 : Sequences
Differentiation Formulas
Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example General Power Fun
Sums and Differences
The Sum Rule if f and g are both differentiable at x, then f + g is
differentiable at x and
d
d
d
[f (x) + g (x)] =
f (x) +
g (x)
dx
dx
dx
The Difference Rule if f and g are both differentiable at x, then f − g is
differentiable at x and
d
d
d
[f (x) − g (x)] =
f (x) −
g (x)
dx
dx
dx
Example
I
Find the derivative of the function f (x) = x 2 + 2x + 4.
f 0 (x) =
Example
d 2
x
dx
+
d
(2x)
dx
+
d
4
dx
= 2x + 2 + 0 = 2x + 2.
Find the derivative of the function f1 (x) = x 12 − 10x 6 + 3x + 1.
Annette Pilkington
Lecture 23 : Sequences
Differentiation Formulas
Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example General Power Fun
Sums and Differences
The Sum Rule if f and g are both differentiable at x, then f + g is
differentiable at x and
d
d
d
[f (x) + g (x)] =
f (x) +
g (x)
dx
dx
dx
The Difference Rule if f and g are both differentiable at x, then f − g is
differentiable at x and
d
d
d
[f (x) − g (x)] =
f (x) −
g (x)
dx
dx
dx
Example
I
Find the derivative of the function f (x) = x 2 + 2x + 4.
f 0 (x) =
Example
I f10 (x)
d 2
x
dx
+
d
(2x)
dx
+
d
4
dx
= 2x + 2 + 0 = 2x + 2.
Find the derivative of the function f1 (x) = x 12 − 10x 6 + 3x + 1.
=
d 12
x
dx
−
d
(10x 6 )
dx
+
d
(3x)
dx
Annette Pilkington
+
d
1
dx
= 12x 11 − 60x 5 + 3 + 0
Lecture 23 : Sequences
Differentiation Formulas
Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example General Power Fun
Sums and Differences
The Sum Rule if f and g are both differentiable at x, then f + g is
differentiable at x and
d
d
d
[f (x) + g (x)] =
f (x) +
g (x)
dx
dx
dx
The Difference Rule if f and g are both differentiable at x, then f − g is
differentiable at x and
d
d
d
[f (x) − g (x)] =
f (x) −
g (x)
dx
dx
dx
Example
I
f 0 (x) =
Example
I f10 (x)
I
Find the derivative of the function f (x) = x 2 + 2x + 4.
d 2
x
dx
+
d
(2x)
dx
+
d
4
dx
= 2x + 2 + 0 = 2x + 2.
Find the derivative of the function f1 (x) = x 12 − 10x 6 + 3x + 1.
=
d 12
x
dx
−
d
(10x 6 )
dx
+
d
(3x)
dx
+
d
1
dx
= 12x 11 − 60x 5 + 3 + 0
= 12x 11 − 60x 5 + 3.
Annette Pilkington
Lecture 23 : Sequences
Differentiation Formulas Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example General Power Fun
The Product Rule
Product Rule: If f and g are both differentiable at x, then f · g is
differentiable at x and
d
d
d
[f (x)g (x)] = f (x) [g (x)] + g (x) [f (x)].
dx
dx
dx
This can be rewritten in a number of ways
d(uv )
du
dv
=v
+ u , or (fg )0 = gf 0 + fg 0 .
dx
dx
dx
Example Let k(x) = x(x 2 + 2x + 4), find k 0 (x).
Example Let F (t) = (t 2 + 4)(2t 3 + t 2 ). Find F 0 (t).
Annette Pilkington
Lecture 23 : Sequences
Differentiation Formulas Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example General Power Fun
The Product Rule
Product Rule: If f and g are both differentiable at x, then f · g is
differentiable at x and
d
d
d
[f (x)g (x)] = f (x) [g (x)] + g (x) [f (x)].
dx
dx
dx
This can be rewritten in a number of ways
d(uv )
du
dv
=v
+ u , or (fg )0 = gf 0 + fg 0 .
dx
dx
dx
Example Let k(x) = x(x 2 + 2x + 4), find k 0 (x).
I
Let f (x) = x and g (x) = x 2 + 2x + 4.
Example Let F (t) = (t 2 + 4)(2t 3 + t 2 ). Find F 0 (t).
Annette Pilkington
Lecture 23 : Sequences
Differentiation Formulas Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example General Power Fun
The Product Rule
Product Rule: If f and g are both differentiable at x, then f · g is
differentiable at x and
d
d
d
[f (x)g (x)] = f (x) [g (x)] + g (x) [f (x)].
dx
dx
dx
This can be rewritten in a number of ways
d(uv )
du
dv
=v
+ u , or (fg )0 = gf 0 + fg 0 .
dx
dx
dx
Example Let k(x) = x(x 2 + 2x + 4), find k 0 (x).
I
Let f (x) = x and g (x) = x 2 + 2x + 4.
I
The product rule says that
2
d
d
d
k(x) = g (x) dx
(f (x) + f (x) dx
g (x) = (x 2 + 2x + 4) dx
+ x d(x +2x+4)
dx
dx
dx
Example Let F (t) = (t 2 + 4)(2t 3 + t 2 ). Find F 0 (t).
Annette Pilkington
Lecture 23 : Sequences
Differentiation Formulas Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example General Power Fun
The Product Rule
Product Rule: If f and g are both differentiable at x, then f · g is
differentiable at x and
d
d
d
[f (x)g (x)] = f (x) [g (x)] + g (x) [f (x)].
dx
dx
dx
This can be rewritten in a number of ways
d(uv )
du
dv
=v
+ u , or (fg )0 = gf 0 + fg 0 .
dx
dx
dx
Example Let k(x) = x(x 2 + 2x + 4), find k 0 (x).
I
Let f (x) = x and g (x) = x 2 + 2x + 4.
I
The product rule says that
2
d
d
d
k(x) = g (x) dx
(f (x) + f (x) dx
g (x) = (x 2 + 2x + 4) dx
+ x d(x +2x+4)
dx
dx
dx
I
= (x 2 + 2x + 4)1 + x(2x + 2) = x 2 + 2x + 4 + 2x 2 + 2x = 3x 2 + 4x + 4.
Example Let F (t) = (t 2 + 4)(2t 3 + t 2 ). Find F 0 (t).
Annette Pilkington
Lecture 23 : Sequences
Differentiation Formulas Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example General Power Fun
The Product Rule
Product Rule: If f and g are both differentiable at x, then f · g is
differentiable at x and
d
d
d
[f (x)g (x)] = f (x) [g (x)] + g (x) [f (x)].
dx
dx
dx
This can be rewritten in a number of ways
d(uv )
du
dv
=v
+ u , or (fg )0 = gf 0 + fg 0 .
dx
dx
dx
Example Let k(x) = x(x 2 + 2x + 4), find k 0 (x).
I
Let f (x) = x and g (x) = x 2 + 2x + 4.
I
The product rule says that
2
d
d
d
k(x) = g (x) dx
(f (x) + f (x) dx
g (x) = (x 2 + 2x + 4) dx
+ x d(x +2x+4)
dx
dx
dx
I
= (x 2 + 2x + 4)1 + x(2x + 2) = x 2 + 2x + 4 + 2x 2 + 2x = 3x 2 + 4x + 4.
Example Let F (t) = (t 2 + 4)(2t 3 + t 2 ). Find F 0 (t).
I
We use the second formula. Let u = t 2 + 4 and let v = 2t 3 + t 2 .
Annette Pilkington
Lecture 23 : Sequences
Differentiation Formulas Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example General Power Fun
The Product Rule
Product Rule: If f and g are both differentiable at x, then f · g is
differentiable at x and
d
d
d
[f (x)g (x)] = f (x) [g (x)] + g (x) [f (x)].
dx
dx
dx
This can be rewritten in a number of ways
d(uv )
du
dv
=v
+ u , or (fg )0 = gf 0 + fg 0 .
dx
dx
dx
Example Let k(x) = x(x 2 + 2x + 4), find k 0 (x).
I
Let f (x) = x and g (x) = x 2 + 2x + 4.
I
The product rule says that
2
d
d
d
k(x) = g (x) dx
(f (x) + f (x) dx
g (x) = (x 2 + 2x + 4) dx
+ x d(x +2x+4)
dx
dx
dx
I
= (x 2 + 2x + 4)1 + x(2x + 2) = x 2 + 2x + 4 + 2x 2 + 2x = 3x 2 + 4x + 4.
Example Let F (t) = (t 2 + 4)(2t 3 + t 2 ). Find F 0 (t).
I
We use the second formula. Let u = t 2 + 4 and let v = 2t 3 + t 2 .
I
Then
d
F (t)
dt
2
3
= v du
+ u dv
= (2t 3 + t 2 ) d(tdt+4) + (t 2 + 4) d(2tdt+t
dt
dt
Annette Pilkington
Lecture 23 : Sequences
2
)
Differentiation Formulas Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example General Power Fun
The Product Rule
Product Rule: If f and g are both differentiable at x, then f · g is
differentiable at x and
d
d
d
[f (x)g (x)] = f (x) [g (x)] + g (x) [f (x)].
dx
dx
dx
This can be rewritten in a number of ways
d(uv )
du
dv
=v
+ u , or (fg )0 = gf 0 + fg 0 .
dx
dx
dx
Example Let k(x) = x(x 2 + 2x + 4), find k 0 (x).
I
Let f (x) = x and g (x) = x 2 + 2x + 4.
I
The product rule says that
2
d
d
d
k(x) = g (x) dx
(f (x) + f (x) dx
g (x) = (x 2 + 2x + 4) dx
+ x d(x +2x+4)
dx
dx
dx
I
= (x 2 + 2x + 4)1 + x(2x + 2) = x 2 + 2x + 4 + 2x 2 + 2x = 3x 2 + 4x + 4.
Example Let F (t) = (t 2 + 4)(2t 3 + t 2 ). Find F 0 (t).
I
We use the second formula. Let u = t 2 + 4 and let v = 2t 3 + t 2 .
I
Then
I
d
F (t)
dt
3
2
2
3
= v du
+ u dv
= (2t 3 + t 2 ) d(tdt+4) + (t 2 + 4) d(2tdt+t
dt
dt
2
2
= (2t + t )(2t) + (t + 4)(6t + 2t)
Annette Pilkington
Lecture 23 : Sequences
2
)
Differentiation Formulas Sums and Differences The Product Rule 14
Special Case of The quotient Rule The Quotient Rule Example General Power Fun
12
Special Case of The quotient Rule
10
d
dx
Let g be differentiable and non-zero at x, then
8
“
1
g (x)
”
0
(x))
= − (gg (x))
2
Example The curve y = 1/(1 + x 2 ) is called a Witch of Maria Agnesi. Find
the equation of the tangent line to the curve at the point (−1, 1/2).
6
4
2
15
– 10
–5
q(x) =
1
1 + x2
5
–2
–4
–6
–8
– 10
– 12
– 14
– 16
Annette Pilkington
Lecture 23 : Sequences
10
15
Differentiation Formulas Sums and Differences The Product Rule 14
Special Case of The quotient Rule The Quotient Rule Example General Power Fun
12
Special Case of The quotient Rule
10
d
dx
Let g be differentiable and non-zero at x, then
8
“
1
g (x)
”
0
(x))
= − (gg (x))
2
Example The curve y = 1/(1 + x 2 ) is called a Witch of Maria Agnesi. Find
the equation of the tangent line to the curve at the point (−1, 1/2).
6
4
2
15
– 10
–5
q(x) =
1
1 + x2
5
–2
–4
I
Let g (x) = x 2 + 1.
–6
–8
– 10
– 12
– 14
– 16
Annette Pilkington
Lecture 23 : Sequences
10
15
Differentiation Formulas Sums and Differences The Product Rule 14
Special Case of The quotient Rule The Quotient Rule Example General Power Fun
12
Special Case of The quotient Rule
10
d
dx
Let g be differentiable and non-zero at x, then
8
“
1
g (x)
”
0
(x))
= − (gg (x))
2
Example The curve y = 1/(1 + x 2 ) is called a Witch of Maria Agnesi. Find
the equation of the tangent line to the curve at the point (−1, 1/2).
6
4
2
15
– 10
q(x) =
–5
1
1 + x2
5
10
–2
–4
I
Let g (x) = x 2 + 1.
I
According to the above rule,
–6
dy
–8
dx
=
−g 0 (x)
(g (x))2
=
−2x
.
(1+x 2 )2
– 10
– 12
– 14
– 16
Annette Pilkington
Lecture 23 : Sequences
15
Differentiation Formulas Sums and Differences The Product Rule 14
Special Case of The quotient Rule The Quotient Rule Example General Power Fun
12
Special Case of The quotient Rule
10
d
dx
Let g be differentiable and non-zero at x, then
8
“
1
g (x)
”
0
(x))
= − (gg (x))
2
Example The curve y = 1/(1 + x 2 ) is called a Witch of Maria Agnesi. Find
the equation of the tangent line to the curve at the point (−1, 1/2).
6
4
2
15
– 10
q(x) =
–5
1
1 + x2
5
10
–2
–4
I
Let g (x) = x 2 + 1.
I
(x)
= −g
=
According to the above rule, dy
dx
(g (x))2
˛
dy ˛
2
2
1
= (1+1)2 = 4 = 2 .
When x = −1, dx ˛
I
–6
0
–8
−2x
.
(1+x 2 )2
– 10
x=−1
– 12
– 14
– 16
Annette Pilkington
Lecture 23 : Sequences
15
Differentiation Formulas Sums and Differences The Product Rule 14
Special Case of The quotient Rule The Quotient Rule Example General Power Fun
12
Special Case of The quotient Rule
10
d
dx
Let g be differentiable and non-zero at x, then
8
“
1
g (x)
”
0
(x))
= − (gg (x))
2
Example The curve y = 1/(1 + x 2 ) is called a Witch of Maria Agnesi. Find
the equation of the tangent line to the curve at the point (−1, 1/2).
6
4
2
15
– 10
q(x) =
–5
1
1 + x2
5
10
15
–2
–4
I
Let g (x) = x 2 + 1.
I
(x)
= −g
=
According to the above rule, dy
dx
(g (x))2
˛
dy ˛
2
2
1
= (1+1)2 = 4 = 2 .
When x = −1, dx ˛
I
–6
0
–8
−2x
.
(1+x 2 )2
– 10
x=−1
I
Since (−1, 1/2) is a point on the tangent, we have the equation of the
tangent to the curve at x = −1 is given by
1
1
1
1
1
y − 1/2 = (x + 1) or
y = x+ =+
or y = x + 1.
2
2
2
2
2
– 12
– 14
– 16
Annette Pilkington
Lecture 23 : Sequences
Differentiation Formulas Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example General Power Fun
The Quotient Rule
We can combine the above rules to get the quotient rule:
If f and g are differentiable at x and g (x) 6= 0, then gf is differentiable at x and
d
d
g (x) dx
[f (x)] − f (x) dx
[g (x)]
d h f (x) i
=
.
2
dx g (x)
[g (x)]
We can rewrite this as
d( vu )
v du − u dv
= dx 2 dx
dx
v
Example
Let K (x) =
x 3 +x 2 +1
,
x 4 +1
or
“ f ”0
g
=
gf 0 − fg 0
.
g2
find K 0 (x). What is K 0 (1)?
Annette Pilkington
Lecture 23 : Sequences
Differentiation Formulas Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example General Power Fun
The Quotient Rule
We can combine the above rules to get the quotient rule:
If f and g are differentiable at x and g (x) 6= 0, then gf is differentiable at x and
d
d
g (x) dx
[f (x)] − f (x) dx
[g (x)]
d h f (x) i
=
.
2
dx g (x)
[g (x)]
We can rewrite this as
d( vu )
v du − u dv
= dx 2 dx
dx
v
Example
I
Let K (x) =
x 3 +x 2 +1
,
x 4 +1
or
“ f ”0
g
=
gf 0 − fg 0
.
g2
find K 0 (x). What is K 0 (1)?
let f (x) = x 3 + x 2 + 1 and g (x) = x 4 + 1.
Annette Pilkington
Lecture 23 : Sequences
Differentiation Formulas Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example General Power Fun
The Quotient Rule
We can combine the above rules to get the quotient rule:
If f and g are differentiable at x and g (x) 6= 0, then gf is differentiable at x and
d
d
g (x) dx
[f (x)] − f (x) dx
[g (x)]
d h f (x) i
=
.
2
dx g (x)
[g (x)]
We can rewrite this as
d( vu )
v du − u dv
= dx 2 dx
dx
v
Example
Let K (x) =
x 3 +x 2 +1
,
x 4 +1
or
“ f ”0
g
gf 0 − fg 0
.
g2
find K 0 (x). What is K 0 (1)?
I
let f (x) = x 3 + x 2 + 1 and g (x) = x 4 + 1.
I
The rule above says
K 0 (x) =
=
d [f (x)]−f (x) d [g (x)]
g (x) dx
dx
[g (x)]2
=
Annette Pilkington
d [x 3 +x 2 +1]−(x 3 +x 2 +1) d (x 4 +1)
(x 4 +1) dx
dx
[(x 4 +1)]2
Lecture 23 : Sequences
Differentiation Formulas Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example General Power Fun
The Quotient Rule
We can combine the above rules to get the quotient rule:
If f and g are differentiable at x and g (x) 6= 0, then gf is differentiable at x and
d
d
g (x) dx
[f (x)] − f (x) dx
[g (x)]
d h f (x) i
=
.
2
dx g (x)
[g (x)]
We can rewrite this as
d( vu )
v du − u dv
= dx 2 dx
dx
v
Example
Let K (x) =
x 3 +x 2 +1
,
x 4 +1
or
“ f ”0
g
let f (x) = x 3 + x 2 + 1 and g (x) = x 4 + 1.
I
The rule above says
I
=
gf 0 − fg 0
.
g2
find K 0 (x). What is K 0 (1)?
I
K 0 (x) =
=
d [f (x)]−f (x) d [g (x)]
g (x) dx
dx
[g (x)]2
=
d [x 3 +x 2 +1]−(x 3 +x 2 +1) d (x 4 +1)
(x 4 +1) dx
dx
[(x 4 +1)]2
(x 4 +1)(3x 2 +2x)−(x 3 +x 2 +1)4x 3
[(x 4 +1)]2
Annette Pilkington
Lecture 23 : Sequences
Differentiation Formulas Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example General Power Fun
The Quotient Rule
We can combine the above rules to get the quotient rule:
If f and g are differentiable at x and g (x) 6= 0, then gf is differentiable at x and
d
d
g (x) dx
[f (x)] − f (x) dx
[g (x)]
d h f (x) i
=
.
2
dx g (x)
[g (x)]
We can rewrite this as
d( vu )
v du − u dv
= dx 2 dx
dx
v
Example
Let K (x) =
x 3 +x 2 +1
,
x 4 +1
or
“ f ”0
g
gf 0 − fg 0
.
g2
find K 0 (x). What is K 0 (1)?
I
let f (x) = x 3 + x 2 + 1 and g (x) = x 4 + 1.
I
The rule above says
K 0 (x) =
=
d [f (x)]−f (x) d [g (x)]
g (x) dx
dx
[g (x)]2
=
d [x 3 +x 2 +1]−(x 3 +x 2 +1) d (x 4 +1)
(x 4 +1) dx
dx
[(x 4 +1)]2
(x 4 +1)(3x 2 +2x)−(x 3 +x 2 +1)4x 3
[(x 4 +1)]2
I
=
I
K 0 (1) =
(2)(5)−(3)4
[(2)]2
=
−2
4
=
−1
2
Annette Pilkington
Lecture 23 : Sequences
Differentiation Formulas Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example General Power Fun
Example
Here’s a way to remember the quotient rule:
”low d-high minus hi d-low, square the bottom and away we go”
or
Low D High minus High D Low, all over the square of what’s below.
Note we should see if we simplify a function with cancellation before we rush
into using the quotient rule.
Example Find the derivative of L(x) =
Annette Pilkington
x 6 +x 4 +x 2
.
x2
Lecture 23 : Sequences
Differentiation Formulas Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example General Power Fun
Example
Here’s a way to remember the quotient rule:
”low d-high minus hi d-low, square the bottom and away we go”
or
Low D High minus High D Low, all over the square of what’s below.
Note we should see if we simplify a function with cancellation before we rush
into using the quotient rule.
x 6 +x 4 +x 2
.
x2
Example Find the derivative of L(x) =
I
L(x) =
x 6 +x 4 +x 2
x2
=
x 2 (x 4 +x 2 +1)
x2
2
x (x
=
Annette Pilkington
4
+x 2 +1)
x2
= x4 + x2 + 1
Lecture 23 : Sequences
Differentiation Formulas Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example General Power Fun
Example
Here’s a way to remember the quotient rule:
”low d-high minus hi d-low, square the bottom and away we go”
or
Low D High minus High D Low, all over the square of what’s below.
Note we should see if we simplify a function with cancellation before we rush
into using the quotient rule.
x 6 +x 4 +x 2
.
x2
Example Find the derivative of L(x) =
x 6 +x 4 +x 2
x2
I
L(x) =
=
I
L0 (x) = 4x 3 + 2x.
x 2 (x 4 +x 2 +1)
x2
2
x (x
=
Annette Pilkington
4
+x 2 +1)
x2
= x4 + x2 + 1
Lecture 23 : Sequences
Differentiation Formulas Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example
General Power Fun
General Power Functions
When n is a positive integer x n = x · x · x · · · · · x, where the product is taken n
times. We define x 0 = 1 and x −n = 1/x n . We can use the quotient rule to
show that. If n is a positive integer
d −n
(x ) = −nx −n−1
dx
In fact it can be shown that if k is any real number
d k
(x ) = kx k−1 .
dx
Example
If H(x) = 2/x 2 + 3/x 3 + 4/x 4 + 1 find H 0 (x).
Example
Find the derivative of f (x) =
√
Annette Pilkington
x+x 2 +x 1/3
√
.
x 2
Lecture 23 : Sequences
Differentiation Formulas Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example
General Power Fun
General Power Functions
When n is a positive integer x n = x · x · x · · · · · x, where the product is taken n
times. We define x 0 = 1 and x −n = 1/x n . We can use the quotient rule to
show that. If n is a positive integer
d −n
(x ) = −nx −n−1
dx
In fact it can be shown that if k is any real number
d k
(x ) = kx k−1 .
dx
Example
I
If H(x) = 2/x 2 + 3/x 3 + 4/x 4 + 1 find H 0 (x).
H(x) = 2x −2 + 3x −3 + 4x −4 + 1
√
Example
Find the derivative of f (x) =
Annette Pilkington
x+x 2 +x 1/3
√
.
x 2
Lecture 23 : Sequences
Differentiation Formulas Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example
General Power Fun
General Power Functions
When n is a positive integer x n = x · x · x · · · · · x, where the product is taken n
times. We define x 0 = 1 and x −n = 1/x n . We can use the quotient rule to
show that. If n is a positive integer
d −n
(x ) = −nx −n−1
dx
In fact it can be shown that if k is any real number
d k
(x ) = kx k−1 .
dx
Example
If H(x) = 2/x 2 + 3/x 3 + 4/x 4 + 1 find H 0 (x).
I
H(x) = 2x −2 + 3x −3 + 4x −4 + 1
I
According to the power rule, H 0 (x) = −4x −3 − 9x −4 − 16x −5
√
Example
Find the derivative of f (x) =
Annette Pilkington
x+x 2 +x 1/3
√
.
x 2
Lecture 23 : Sequences
Differentiation Formulas Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example
General Power Fun
General Power Functions
When n is a positive integer x n = x · x · x · · · · · x, where the product is taken n
times. We define x 0 = 1 and x −n = 1/x n . We can use the quotient rule to
show that. If n is a positive integer
d −n
(x ) = −nx −n−1
dx
In fact it can be shown that if k is any real number
d k
(x ) = kx k−1 .
dx
Example
If H(x) = 2/x 2 + 3/x 3 + 4/x 4 + 1 find H 0 (x).
I
H(x) = 2x −2 + 3x −3 + 4x −4 + 1
I
According to the power rule, H 0 (x) = −4x −3 − 9x −4 − 16x −5
√
Example
I
Find the derivative of f (x) =
x+x 2 +x 1/3
√
.
x 2
We
can use the quotient rule, or divide each term in the numerator by
√
x 2.
Annette Pilkington
Lecture 23 : Sequences
Differentiation Formulas Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example
General Power Fun
General Power Functions
When n is a positive integer x n = x · x · x · · · · · x, where the product is taken n
times. We define x 0 = 1 and x −n = 1/x n . We can use the quotient rule to
show that. If n is a positive integer
d −n
(x ) = −nx −n−1
dx
In fact it can be shown that if k is any real number
d k
(x ) = kx k−1 .
dx
Example
If H(x) = 2/x 2 + 3/x 3 + 4/x 4 + 1 find H 0 (x).
I
H(x) = 2x −2 + 3x −3 + 4x −4 + 1
I
According to the power rule, H 0 (x) = −4x −3 − 9x −4 − 16x −5
√
Example
Find the derivative of f (x) =
x+x 2 +x 1/3
√
.
x 2
I
We
can use the quotient rule, or divide each term in the numerator by
√
x 2.
I
Dividing by x
√
2
1
√
2)
√
, we get f (x) = x ( 2 −
+ x (2−
Annette Pilkington
Lecture 23 : Sequences
2)
1
√
2)
+ x( 3 −
.
Differentiation Formulas Sums and Differences The Product Rule Special Case of The quotient Rule The Quotient Rule Example
General Power Fun
General Power Functions
When n is a positive integer x n = x · x · x · · · · · x, where the product is taken n
times. We define x 0 = 1 and x −n = 1/x n . We can use the quotient rule to
show that. If n is a positive integer
d −n
(x ) = −nx −n−1
dx
In fact it can be shown that if k is any real number
d k
(x ) = kx k−1 .
dx
Example
If H(x) = 2/x 2 + 3/x 3 + 4/x 4 + 1 find H 0 (x).
I
H(x) = 2x −2 + 3x −3 + 4x −4 + 1
I
According to the power rule, H 0 (x) = −4x −3 − 9x −4 − 16x −5
√
Example
Find the derivative of f (x) =
x+x 2 +x 1/3
√
.
x 2
I
We
can use the quotient rule, or divide each term in the numerator by
√
x 2.
I
Dividing by x
√
I
1
√
√
1
√
, we get f (x) = x ( 2 − 2) + x (2− 2) + x ( 3 − 2) .
√
√
√
√
√
√
1
2
f 0 (x) = ( 21 − 2)x (− 2 − 2) + (2 − 2)x (1− 2) + ( 13 − 2)x (− 3 − 2)
2
Annette Pilkington
Lecture 23 : Sequences