Limit of a derivative - SOLUTIONS
The following problems require the use of the limit definition of a derivative, which is given by
.
They range in difficulty from easy to somewhat challenging. If you are going to try these problems
before looking at the solutions, you can avoid common mistakes by making proper use of functional
notation and careful use of basic algebra. Keep in mind that the goal (in most cases) of these types of
problems is to be able to divide out the
term so that the indeterminant form
can be circumvented and the limit can be calculated.
PROBLEM 1: Use the limit definition to compute the derivative, f'(x), for
.
(Algebraically and arithmetically simplify the expression in the numerator.)
(The term
now divides out and the limit can be calculated.)
.
of the expression
PROBLEM 2 : Use the limit definition to compute the derivative, f'(x), for
.
(Algebraically and arithmetically simplify the expression in the numerator.)
(Factor
(The term
from the expression in the numerator.)
now divides out and the limit can be calculated.)
.
PROBLEM 3 : Use the limit definition to compute the derivative, f'(x), for
.
(Eliminate the square root terms in the numerator of the expression by multiplying
by the conjugate of the numerator divided by itself.)
(Recall that
(The term
)
now divides out and the limit can be calculated.)
.
PROBLEM 4 : Use the limit definition to compute the derivative, f'(x), for
.
(Get a common denominator for the expression in the numerator. Recall that division by
the same as multiplication by
.)
is
(Algebraically and arithmetically simplify the expression in the numerator. It is important to note
that the denominator of this expression should be left in factored form so that the term
can be
easily eliminated later.)
(The term
now divides out and the limit can be calculated.)
.
PROBLEM 5 : Use the limit definition to compute the derivative, f'(x), for
.
(At this point it may appear that multiplying by the conjugate of the numerator over
itself is a good next step. However, doing something else is a better idea.)
(Note that A - B can be written as the difference of cubes , so that
. This will help explain the
next step.)
(Algebraically and arithmetically simplify the expression in the numerator.)
(The term
now divides out and the limit can be calculated.)
.
PROBLEM 6 : Use the limit definition to compute the derivative, f'(x), for
.
(Recall a well-known trigonometry identity :
.)
(Recall the following two well-known trigonometry limits :
and
.)
.
Chain Rule
The following problems require the use of the chain rule. The chain rule is a rule for differentiating
compositions of functions. In the following discussion and solutions the derivative of a function h(x)
will be denoted by
or h'(x) .
.
However, we rarely use this formal approach when applying the chain rule to specific problems.
Instead, we invoke an intuitive approach. For example, it is sometimes easier to think of the functions
f and g as ``layers'' of a problem. Function f is the ``outer layer'' and function g is the ``inner layer.''
Thus, the chain rule tells us to first differentiate the outer layer, leaving the inner layer unchanged
(the term f'( g(x) ) ) , then differentiate the inner layer (the term g'(x) ) . This process will become
clearer as you do the problems. In most cases, final answers are given in the most simplified form.
PROBLEM 1 : Differentiate
.
= 2 (3x+1) (3)
= 6 (3x+1) .
PROBLEM 2 : Differentiate
.
( The outer layer is ``the square root'' and the inner layer is
square root'' first, leaving
Thus,
unchanged. Then differentiate
. Differentiate ``the
.)
PROBLEM 3 : Differentiate
.
( The outer layer is ``the 30th power'' and the inner layer is
30th power'' first, leaving
4 : Differentiate
unchanged. Then differentiate
. ) Thus,
.
( The outer layer is ``the one-third power'' and the inner layer is
one-third power'' first, leaving
. Differentiate ``the
unchanged. Then differentiate
. Differentiate ``the
. ) Thus,
(At this point, we will continue to simplify the expression, leaving the final answer with no
negative exponents.)
.
SOLUTION 5 : Differentiate
.
( First, begin by simplifying the expression before we differentiate it. ) Thus,
( The outer layer is ``the negative four-fifths power'' and the inner layer is
Differentiate ``the negative four-fifths power'' first, leaving
differentiate
.
unchanged. Then
.)
(At this point, we will continue to simplify the expression, leaving the final answer with no
negative exponents.)
.
SOLUTION 6 : Differentiate
.
( The outer layer is ``the sine function'' and the inner layer is (5x) . Differentiate ``the sine
function'' first, leaving (5x) unchanged. Then differentiate (5x) . ) Thus,
.
SOLUTION 7 : Differentiate
.
( The outer layer is ``the exponential function'' and the inner layer is
. Differentiate ``the exponential function'' first, leaving
Then differentiate
. ) Thus,
.
. Recall that
unchanged.
SOLUTION 8 : Differentiate
.
( The outer layer is ``2 raised to a power'' and the inner layer is
differentiate
. Recall that
. Differentiate ``2 raised to a power'' first, leaving
. ) Thus,
unchanged. Then
.
SOLUTION 9 : Differentiate
.
( Since 3 is a MULTIPLIED CONSTANT, we will first use the rule
, where
c is a constant . Hence, the constant 3 just ``tags along'' during the differentiation process. It is
NOT necessary to use the product rule. ) Thus,
( Now the outer layer is ``the tangent function'' and the inner layer is
tangent function'' first, leaving
unchanged. Then differentiate
.
.)
. Differentiate ``the
SOLUTION 10 : Differentiate
.
( The outer layer is ``the natural logarithm (base e) function'' and the inner layer is ( 17-x ) .
Recall that
. Differentiate ``the natural logarithm function'' first, leaving ( 17-x )
unchanged. Then differentiate ( 17-x ). ) Thus,
.