Introduction to Derivatives
Techniques of Differentiation
Basic Rules
Because the derivative is a limit, many of the rules of limits apply to the derivative:
- (cf (x))' = c(f'(x)) where c is a constant. This says that the derivative of a scalar multiple of a function is equal to the derivative of the function multiplied by the scalar multiple.
- (f (x) + g(x)) = f'(x) + g'(x) . The derivative of a sum of two functions is equal to the sum of the individual derivatives.
The Power Rule
This is a powerful way of finding the derivative of a polynomial function. It says:
x
n
 = nx
n-1
|
where n is a real number. For example,
x
4
 = 4x
3
|
The Product Rule
If f and g are two differentiable functions, then (fg)' = f'g + g'f . For example,
 (3x)(
 = 3 +3x(
x
-
)
|
The Quotient Rule
If f and g are two differentiable functions, then
 = 
|
For example,
 =  = 
|
Trigonometric derivatives
The basic trigonometric functions have derivatives that should be memorized: If x is expressed in radians, then:
| (sin(x))' | = cos(x) | ||
| (cos(x))' | = - sin(x) | ||
| (tan(x))' | = sec2(x) = 
|
The Chain Rule
This is a rule for evaluating the derivatives of composite functions
f
o
g
|
= 
f'(g(x)
g'(x)
|
||
| or | |||
| (f (g(x))' | =
f'(g(x)
g'(x)
|
For example, the function f (x) = (3x + 2)2 is a composite function where the outer function, f , is a power function ( u 2 ), and the inner function, g , is a linear function ( 3x + 2 ).
To differentiate this composite function, first treat the inner function as a single variable, and take the derivative of the outer function. Then multiply by the derivative of the inner function:
 3x+2 = 23x+2 (3)
|
Implicit Differentiation
This is a means of finding
, the derivative of
y
with respect to
x
,
even when we do not have a function of the form
y = f (x)
.
Example: Find the slope of the graph at
(0, 0)
for the following function:
| xy 2 = x + y |
To solve this problem, we must essentially first find
and then plug in the
point (0,0) to find the slope.
One possibility would be to rearrange the equation to solve for y explicitly and then use standard procedures to find the derivative. However, this is not always possible, and is indeed difficult to do here.
Instead, the procedure of implicit differentiation involves taking the derivative of both sides of the equation with respect to x . Remember that y is a function of x , so taking the derivative of a function of y involves using the chain rule:
| xy 2 = x + y |
Differentiate both sides with respect to x First consider xy 2 . x and y are both functions of x , so the product rule is needed here. Also, the chain rule is needed to evaluate the derivative of y 2 with respect to x :
xy
2
|
= (1)
y
2
 + x
 2y
|
||
|
(x + y) |
= 1 + 1
|
Recombining both sides of the equation yields:
|
(1)
y
2
+ x
2y
= 1 + 1
|
Now, solving algebraically for
yields
= 
|
Finally, the slope at (0,0) is
=  = - 1
|
