Product Rule
Quotient Rule
Derivative of f(x) /g(x) equals
(f '(x) • g(x)) - (f(x) • g'(x))
g²(x)
Derivative of f(x)g(x) = (f '(x) g(x)) + (f(x)g'(x))
• EXAMPLE : The derivative of
EXAMPLE : The derivative of
(x³ + 5x² -6x + 9) • (7x³ -x² -8x + 1)
(5x² + 2x + 9)
(7x² -3x + 8)
Equals
equals
(3x² + 10x -6)(7x³ - x² -8x + 1) + (x³ + 5x² -6x + 9)(21x² -2x 8)
(10x + 2) • (7x² -3x + 8) - (5x² + 2x + 9) • (14x -3)
(7x² -3x + 8)²
Chain Rule
• First, we should discuss the concept of the composition
of a function which actually means the function of
another function. It is easier to discuss this concept in
informal terms.
ALL compositions of 2 functions consist of 2 parts:
1) The function inside the parentheses and
2) The function outside of the parentheses.
• As an example, let's analyze 4•(x³+5)²
Speaking informally we could say the "inside function" is
(x³+5) and
the "outside function" is 4•(inside) ².
First, let's multiply this out and then take the derivative.
4•(x³+5) ² = 4x6 + 40 x³ + 100
derivative = 24x5 + 120 x²
•
Now, let's differentiate the same equation using the chain rule which
states that the
derivative of a composite function equals:
(derivative of outside) • (inside) • (derivative of inside).
•
•
Using the chain rule to differentiate 4(x3+5)2 we obtain:
derivative of outside = 4 • 2 = 8
inside = x3 + 5
derivative of inside = 3x 2
Now we multiply all 3 quantities to obtain:
ANSWER = 8(x3+5)(3x 2)
As a double check we multiply this out and obtain:
8x3+40 • (3x2) = 24 x5 + 120 x2 which is precisely the answer we
obtained by using the "long way".
What is the derivative of sin(5x3 + 2x) ?
Newton – Raphson Method
Differentiate
Dr. Noreen Quinn
Differentiate
1
Newton’s Method
f ( x) =
1.5
1 2
Finding a root for: f ( x ) = x − 3
2
1 2
x −3
2
f ′( x) = x
5
1.5
4
3
1
f ( 3) = ⋅ 32 − 3 = 1.5
2
Guess:
3
We will use Newton’s
Method to find the
root between 2 and 3.
2
1
-4
-3
-2
-1
0
1
2
3
z 3
2
mtangent = f ′ ( 3) = 3
(not drawn to scale)
4
-1
1.5
3=
1.5
z
3−
= 2.5
(new guess)
3
-2
-3
z=
1.5
3
→
f ( x) =
1.5
→
1 2
x −3
2
f ( x) =
1.5
f ′( x) = x
f ′( x) = x
2.5
1
2
f ( 2.5 ) = ( 2.5 ) − 3 = .125
2
Guess: 2.45
Guess:
z
2
3
f ( 2.45 ) = .00125
z
2
3
mtangent = f ′ ( 2.5 ) = 2.5
2.5 −
.125
= 2.45
2.5 (new guess)
z=
mtangent = f ′ ( 2.45 ) = 2.45
.125
2.5
z=
→
Guess:
1 2
x −3
2
2.44948979592
2.45 −
.00125
= 2.44948979592
2.45
Guess:
.00125
2.45
(new guess)
→
2.44948979592
f ( xn )
f ( 2.44948979592
) = .00000013016
xn +1 = xn −
Newton’s Method:
f ′ ( xn )
f ( 2.44948979592 ) = .00000013016
Amazingly close to zero!
Amazingly close to zero!
This is Newton’s Method of finding roots. It is an example
of an algorithm (a specific set of computational steps.)
This is Newton’s Method of finding roots. It is an example
of an algorithm (a specific set of computational steps.)
It is sometimes called the Newton-Raphson method
It is sometimes called the Newton-Raphson method
This is a recursive algorithm because a set of steps are
repeated with the previous answer put in the next
repetition. Each repetition is called an iteration.
This is a recursive algorithm because a set of steps are
repeated with the previous answer put in the next
repetition. Each repetition is called an iteration.
→
→
2
Find the roots of f ( x ) = x3 − x − 1
Derivation (on board)
Where
x0=1
f ′ ( x ) = 3x 2 − 1
n
xn
f ( xn )
0
1
−1
2
1
1.5
.875
5.75
xn +1 = xn −
f ′ ( xn )
1−
1.5 −
2 1.3478261 .1006822 4.4499055
(1.3252004 )
3
f ( xn )
f ′ ( xn )
−1
= 1.5
2
.875
= 1.3478261
5.75
1.3252004
− 1.3252004 = 1.0020584
≈1
→
Newton-Raphson
There are some limitations to Newton’s method:
Taking 1 as the first approximation of a root
of x3 + 2x − 4 = 0, use the NewtonRaphson method to calculate the root of
this function to 3 decimal places.
Looking for this root.
Bad guess.
Wrong root found
Failure to converge
→
3