Download Limits to Derivatives

Limits to Derivatives
Remember the last class? Good! I believe I mentioned that it was a rather important class.
Let’s review:
There are two ways to use limits to find the slope of a tangent.
Method 1:
mtangent  lim
xa
f ( x)  f (a )
xa
Method 2:
mtangent  lim
h 0
f ( a  h)  f ( a )
h
This is the whole basis to calculus! We can use this equation to find the slope for any
equation.
Example: Find the slope of the curve f(x)=x2-7x+1 at the point (-2,19).
a  2, f (a )  19
f ( a  h)  f ( a )
h 0
h
f (h  2)  19
 lim
h 0
h
(h  2)2  7(h  2)  1 19
 lim
h 0
h
2
h  4h  4  7 h  14  1  19
 lim
h 0
h
2
h  11h
 lim
h 0
h
h(h  11)
 lim
h 0
h
 lim(h  11)
mtangent  lim
mtangent
mtangent
mtangent
mtangent
mtangent
mtangent
h 0
mtangent  11
Easy….right?
Example: Find the slope of the curve f(x)=x2 at the point (-3,9).
a  3, f (a )  9
f ( a  h)  f ( a )
h 0
h
f (h  3)  9
 lim
h 0
h
(h  3) 2  9

 lim
h 0
h
2
h  6h  9  9
 lim
h 0
h
2
h  6h
 lim
h 0
h
h(h  6)
 lim
h 0
h
 lim(h  6)
mtangent  lim
mtangent
mtangent
mtangent
mtangent
mtangent
mtangent
h 0
mtangent  6
Now we can generalize this for any point.
Find the slope of the curve f(x)=x2 at the point (x,y).
a  x, f ( a )  y  x 2
f ( a  h)  f ( a )
h 0
h
f (h  x)  y
 lim
h 0
h
(h  x)2  x 2
 lim
h 0
h
2
h  2 xh  x 2  x 2
 lim
h 0
h
2
h  2 xh
 lim
h 0
h
h( h  2 x )
 lim
h 0
h
 lim(h  2 x)
mtangent  lim
mtangent
mtangent
mtangent
mtangent
mtangent
mtangent
h 0
mtangent  2 x
Finding the general rule to describe the functions at any point is called finding the
derivative. The derivative is a function that represents the slope of a function at any
point. We use two main ways to represent a function.
The derivative of a function f(x) is f’(x) [we say: “f prime of x” or simply “the derivative
of f”].
df (x)
dy
Another way to represent the derivative of the relation f (x)  y is
or
[we
dx
dx
say: “the derivative of f”].
Therefore in the problem above we can say:
The derivative of f(x)=x2 is
f’(x)=2x
or
dy
 2x .
dx
We now have a general solution for the derivative of f(x)=x2. If we were to do this for
other simple polynomials, we may get a pattern.
Take a look at this pattern
f ( x)  c
f '( x)  0
f ( x)  x
f '( x)  1
f ( x)  x 2
f '( x)  2 x
f ( x)  x3
f '( x)  3x 2
f ( x)  x 4
f '( x)  4 x3
f ( x)  x5 f '( x)  5 x 4
hmmmm…..there’s a pattern it seems.
(i) f ( x)  x n
f '( x)  nx n1
lim cf ( x)  c lim f ( x)
Now if you look back at the rules on limits:
x a
x a
lim  f ( x)  g ( x)   lim f ( x)  lim g ( x)
x a
We can apply them to develop a few more general rules.
(ii) f ( x)  ax n
f '( x)  anx n1
Furthermore,
(iii) f ( x)  ax n  bx m
f '( x)  anx n1  bmx m1
These are some of the most useful rules in calculus this year.
You should also remember that
(iv) f ( x)  c
f '( x)  0
This is true since f ( x)  c  cx0
f '( x)  c  0 x01  0
x a
x a
Let’s use these rules:
Example 1:(a) Find the derivative of f ( x)  5 x3  4 x 2  7 x  5
Solution: f '( x)  15x 2  8x  7
(b) Find the slope of the tangent of f ( x ) at the point (2,33).
Solution: f '(2)  15(2)2  8(2)  7  51
So the slope when x=2 of the function is 51.
Easy
Example 2:(a) Find the derivative of f ( x)  x4  3x2  8x  3
Solution: f '( x)  4 x3  6 x  8
(b) Find the slope of the tangent of f ( x ) when x=1.
Solution: f '(1)  4(1)3  6(1)  8  6
So the slope when x=1 of the function is -6.
Easy (again)
Notice that this rule would work for any function that can be written as an exponent.
Example 3:(a) Find the derivative of f ( x)  x
1
Solution: f ( x)  x 2
1 12 1 1  12
f '( x)  x  x
2
2
This could be rewritten as
1
1
x
f '( x)  1 

2 x 2x
2x 2
(but that’s not critical)
(b) Find the slope of the tangent of f ( x ) when x=9.
1
Solution: f '(9) 
6
1
So the slope when x=9 of the function is .
6
Easy (as usual)
Example 4:(a) Find the derivative of f ( x ) 
1
x
Solution: f ( x)  x 1
f '( x)   x 11   x2
This could be rewritten as
f '( x)  
1
x2
(b) Find the slope of the tangent of f ( x ) when x=2.
1
Solution: f '(2) 
4
1
So the slope when x=9 of the function is .
4
Easy (this is almost getting boring)
Example 5:(a) Find the derivative of f ( x)  4 x 2  6 x 
1
Solution: f ( x)  4 x2  6 x 2  3x2

1
f '( x)  8x  3x 2  6 x3
This could be rewritten as
3 x 6
f '( x)  8 x 
 3
x
x
Easy (Yawn)
3
x2