Basic Derivative Rules

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Derivative Rules: Evaluation Rules | Practice Problems | Quiz

NOTE: The examples and practice problems on this page make use of the differentiation rules for polynomials, trigonometric, and exponential functions.

EVALUATION RULES: (top)

1. Constant Rule

The derivative of a constant is zero; that is, for a constant c:

Explanation

PROBLEM 1:
Find the derivative of the function .
Solution

This function is simply a constant, equal to approximately 5.2; thus, it's derivative is 0.

2. Constant Multiple Rule

The derivative of a constant multiplied by a function is the constant multiplied by the derivative of the original function:

.

Explanation

This is easily proved using the definition of derivative. For clarity, let g(x)=af(x):

PROBLEM 2:
Evaluate .

Solution

Always be able to justify each step when evaluating a derivative.

by the constant multiple rule

by evaluating the derivative with the power rule and the constant rule

by simplification

3. Multiple Rule

The derivative of f(ax) is given by

Explanation

This rule is a special case of the chain rule.

PROBLEM 3:
Find the derivative of 3sin(3x).

Solution

We'll use the multiple rule:

=
by the constant multiple rule

=
by the multiple rule

=
since 3*3=9 and the derivative of sine is cosine

4. Sum/Difference Rules

The derivative of the sum of two functions is the sum of the derivatives of the two functions:

.

Likewise, the derivative of the difference of two functions is the difference of the derivatives of the two functions.

Explanation

This rule, too, is easily proved from the definition of derivative. Again for clarity, let j(x)=f(x)+g(x):

PROBLEM 4:
Find the derivative of .

Solution

Evaluate the derivative step-by-step using the derivative rules.

=
by the sum and difference rules

=
by the constant multiple rule

=
by evaluating the derivatives of x², 3x, and tan(x).

5. Product Rule

The derivative of the product of two functions is NOT the product of the functions' derivatives; rather, it is described by the equation below:

.

Explanation

=
by the definition of derivative

=
by the definition of j(x)

Now comes the trick: we'll both add and subtract in the numerator.
Note that this does not change the equation as the net addition is 0. The result is

=
by rearrangement

=
by factoring

=
by evaluation of the limit and definition of derivative for f(x) and g(x)

PROBLEM 5:
Find the derivative of .

Solution

Evaluate the derivative step-by-step using the derivative rules.

=
by the product rule

=
by evaluating the derivatives with the power and constant rules

=
by simplifying

6. Quotient Rule

The derivative of the quotient of two functions is NOT the quotient of the functions' derivatives; rather, it is described by the equation below:

.

Explanation

We could, of course, prove this rule from the definition of derivative, but it's much easier to simply use the product rule, which we just proved, and some algebraic manipulation:

First, let
.

<=>
by multiplying both sides of the equation by g(x)

<=>
by differentiating f(x) using the product rule

Our goal is then to isolate h'(x):

(simple algebraic maniuplation)

<=>
(simple algebraic maniuplation)

<=>
by multiplying through by g(x) to get rid of the fraction in the numerator.

PROBLEM 6:
Evaluate the derivative of
Solution

=
by the quotient rule

=
by evaluating the derivatives

=
simplification

=
simplification

simplification

7. Chain Rule

The chain rule is used to differentiate composite functions. As such, it is a vital tool for differentiating most functions of a certain complexity. It states:
.

Explanation

The proof of the chain rule is somewhat more difficult than the other rules developed on this page. As usual, we'll start with the definition of derivative. Note that the definition of derivative used here uses slightly different notation than in previous proofs but is formally identical.

=
(definition of derivative)

We'll start out by presenting a proof that is fairly easy to understand but contains a small technical error, then explain and correct the error.

The first thing to do is multiply the top and bottom of the definition of derivative by 1 in a clever way:

=
(multiplying by 1)

=
by rearranging terms and separating the limit in two

Note that the second limit is the definition of derivative for g(x). Now we need to evaluate the first limit. Since we've assumed g is differentiable, g is also continuous. This means that as . Consequently, we can rewrite the first limit as

=
by the definition of derivative

Thus, it seems we have shown that .

There is, however, a flaw in the proof as above: namely, we can't be sure that , and if it were, our first step would have been to multiply and divide by 0, which is not allowed.

To get around this problem, we're going to define a function very similar to but just different enough to avoid the flaw in the proof above:

Let
=

From the definition of and the discussion above, we can see that (*).

Now, all we need to do is show that = so that we can use right-hand side of this equation instead of the definition of derivative (which is the left-hand side).

There are two possible cases here. If , then by the definition of ,

If, on the other hand, g(x₀) = g(x), then =0 and =0,
so = reduces to the true statement 0=0.

Thus, = (**), so

=
by the definition of derivative

=
by **

=
by * and the definition of derivative for g(x)

This completes the proof of the chain rule: that .

PROBLEM 7:
Evaluate .
Solution

We'll regard the function to be differentiated as the composition of two functions f and g, where

and . Then

=
by the constant multiple rule

=
by the chain rule

=
6(x² + 2)(2x) since and

=
12(x³ + 2x) simplification

PRACTICE PROBLEMS: (top)

Evaluate the derivatives of the following:

Solution

	=		by the constant multiple rule
	=		by the product rule
	=		by evaluating the derivatives using the power rule, the chain rule, and the exponential rule
	=		by factoring

Solution

	=		by the chain rule
	=		by the chain rule and the exponential rule
	=		by the sum rule
	=		by the chain rule
	=		simplification

Solution

	=		by the quotient rule
	=		by the product and chain rules in the numerator
	=		by evaluating the simple derivatives
	=		simplification
	=		by factoring
	=		simplification

QUIZ: (top)

This page last updated 6 August, 2008 5:57 PM

		by the constant multiple rule
		by evaluating the derivative with the power rule and the constant rule
		by simplification

	=		by the constant multiple rule
	=		by the multiple rule
	=		since 3*3=9 and the derivative of sine is cosine

	=		by the sum and difference rules
	=		by the constant multiple rule
	=		by evaluating the derivatives of x², 3x, and tan(x).

	=	by the definition of derivative
	=	by the definition of j(x)
Now comes the trick: we'll both add and subtract in the numerator. Note that this does not change the equation as the net addition is 0. The result is

=		by rearrangement
=		by factoring
=		by evaluation of the limit and definition of derivative for f(x) and g(x)

	=		by the product rule
	=		by evaluating the derivatives with the power and constant rules
	=		by simplifying

First, let	.
<=>		by multiplying both sides of the equation by g(x)
<=>		by differentiating f(x) using the product rule
Our goal is then to isolate h'(x):
		(simple algebraic maniuplation)
<=>		(simple algebraic maniuplation)
<=>		by multiplying through by g(x) to get rid of the fraction in the numerator.

	=		by the quotient rule
	=		by evaluating the derivatives
	=		simplification
	=		simplification
			simplification

The proof of the chain rule is somewhat more difficult than the other rules developed on this page. As usual, we'll start with the definition of derivative. Note that the definition of derivative used here uses slightly different notation than in previous proofs but is formally identical.
	=	(definition of derivative)
We'll start out by presenting a proof that is fairly easy to understand but contains a small technical error, then explain and correct the error. The first thing to do is multiply the top and bottom of the definition of derivative by 1 in a clever way:
	=	(multiplying by 1)
	=	by rearranging terms and separating the limit in two
Note that the second limit is the definition of derivative for g(x). Now we need to evaluate the first limit. Since we've assumed g is differentiable, g is also continuous. This means that as . Consequently, we can rewrite the first limit as

	=	by the definition of derivative
Thus, it seems we have shown that . There is, however, a flaw in the proof as above: namely, we can't be sure that , and if it were, our first step would have been to multiply and divide by 0, which is not allowed. To get around this problem, we're going to define a function very similar to but just different enough to avoid the flaw in the proof above:
Let	=
From the definition of and the discussion above, we can see that (*). Now, all we need to do is show that = so that we can use right-hand side of this equation instead of the definition of derivative (which is the left-hand side).
There are two possible cases here. If , then by the definition of ,

If, on the other hand, g(x₀) = g(x), then =0 and =0, so = reduces to the true statement 0=0.
Thus, = (**), so
	=	by the definition of derivative
	=	by **
	=	by * and the definition of derivative for g(x)
This completes the proof of the chain rule: that .

We'll regard the function to be differentiated as the composition of two functions f and g, where
and . Then
	=		by the constant multiple rule
	=		by the chain rule
	=	6(x² + 2)(2x)	since and
	=	12(x³ + 2x)	simplification