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

This is easily proved using the definition of derivative. For clarity, let g(x)=af(x):
 Solution
 Always be able to justify each step when evaluating a derivative.
by the constant multiple ruleby evaluating the derivative with the power rule and the constant ruleby simplification  Explanation
 This rule is a special case of the chain rule.
 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  Explanation

This rule, too, is easily proved from the definition of derivative. Again for clarity, let j(x)=f(x)+g(x):
 Solution
 Evaluate the derivative stepbystep using the derivative rules.
=by the sum and difference rules =by the constant multiple rule =by evaluating the derivatives of x^{2}, 3x, and tan(x).  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)  Solution
 Evaluate the derivative stepbystep using the derivative rules.
=by the product rule =by evaluating the derivatives with the power and constant rules =by simplifying  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.  Solution
 =
by the quotient rule =by evaluating the derivatives =simplification =simplification simplification  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 righthand side of this equation instead of the definition of derivative (which is the lefthand side).
There are two possible cases here. If , then by the definition of , If, on the other hand, g(x_{0}) = 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 .  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(x^{2} + 2)(2x) since and =12(x^{3} + 2x) simplification 
 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
Basic Derivative Rules
Home > Calculus 1 > Derivative Rules
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)
The derivative of a constant is zero; that is, for a constant c:
PROBLEM 1:
Find the derivative of the function .
The derivative of a constant multiplied by a function is the constant multiplied by the derivative of the original function:
.
PROBLEM 2:
Evaluate .
The derivative of f(ax) is given by
PROBLEM 3:
Find the derivative of 3sin(3x).
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.
PROBLEM 4:
Find the derivative of .
The derivative of the product of two functions is NOT the product of the functions' derivatives; rather, it is described by the equation below:
.
PROBLEM 5:
Find the derivative of .
The derivative of the quotient of two functions is NOT the quotient of the functions' derivatives; rather, it is described by the equation below:
.
PROBLEM 6:7. Chain Rule
Evaluate the derivative of
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:.
PROBLEM 7:
Evaluate .
PRACTICE PROBLEMS: (top)
Evaluate the derivatives of the following:QUIZ: (top)
This page last updated 6 August, 2008 5:57 PM