Limit of a derivative - SOLUTIONS
Limit of a derivative - SOLUTIONS
The following problems require the use of the limit definition of a derivative, which is given by
[pic].
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 [pic]term so that the indeterminant form [pic]of the expression can be circumvented and the limit can be calculated.
PROBLEM 1: Use the limit definition to compute the derivative, f'(x), for
. [pic]
[pic]
[pic]
(Algebraically and arithmetically simplify the expression in the numerator.)
[pic]
(The term [pic]now divides out and the limit can be calculated.)
[pic]
[pic].
PROBLEM 2 : Use the limit definition to compute the derivative, f'(x), for
[pic]. [pic]
[pic]
(Algebraically and arithmetically simplify the expression in the numerator.)
[pic]
[pic]
(Factor [pic]from the expression in the numerator.)
[pic]
(The term [pic]now divides out and the limit can be calculated.)
[pic]
[pic].
PROBLEM 3 : Use the limit definition to compute the derivative, f'(x), for
[pic].
[pic]
[pic]
[pic]
(Eliminate the square root terms in the numerator of the expression by multiplying
by the conjugate of the numerator divided by itself.)
[pic]
(Recall that [pic])
[pic]
[pic]
(The term [pic]now divides out and the limit can be calculated.)
[pic]
[pic]
[pic].
• PROBLEM 4 : Use the limit definition to compute the derivative, f'(x), for
[pic]. [pic]
[pic]
(Get a common denominator for the expression in the numerator. Recall that division by [pic]is the same as multiplication by [pic]. )
[pic]
(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 [pic]can be easily eliminated later.)
[pic]
[pic]
(The term [pic]now divides out and the limit can be calculated.)
[pic]
[pic]
[pic].
• PROBLEM 5 : Use the limit definition to compute the derivative, f'(x), for
[pic]. [pic]
[pic]
(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.)
[pic]
(Note that A - B can be written as the difference of cubes , so that
[pic]. This will help explain the next step.)
[pic]
[pic]
(Algebraically and arithmetically simplify the expression in the numerator.)
[pic]
[pic]
(The term [pic]now divides out and the limit can be calculated.)
[pic]
[pic]
[pic]
[pic].
PROBLEM 6 : Use the limit definition to compute the derivative, f'(x), for
[pic]. [pic]
[pic]
[pic]
(Recall a well-known trigonometry identity :
[pic].)
[pic]
[pic]
(Recall the following two well-known trigonometry limits :
[pic]and [pic].)
[pic]
[pic]
[pic]
[pic]
[pic].
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 [pic]or h'(x) .
[pic].
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 [pic].
[pic]
= 2 (3x+1) (3)
= 6 (3x+1) .
PROBLEM 2 : Differentiate [pic].
( The outer layer is ``the square root'' and the inner layer is [pic]. Differentiate ``the square root'' first, leaving [pic]unchanged. Then differentiate [pic]. ) Thus,
[pic]
[pic]
[pic]
[pic]
[pic]
PROBLEM 3 : Differentiate [pic].
( The outer layer is ``the 30th power'' and the inner layer is [pic]. Differentiate ``the 30th power'' first, leaving [pic]unchanged. Then differentiate [pic]. ) Thus,
[pic]
[pic]
[pic]
4 : Differentiate [pic].
( The outer layer is ``the one-third power'' and the inner layer is [pic]. Differentiate ``the one-third power'' first, leaving [pic]unchanged. Then differentiate [pic]. ) Thus,
[pic]
[pic]
(At this point, we will continue to simplify the expression, leaving the final answer with no negative exponents.)
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic].
SOLUTION 5 : Differentiate [pic].
( First, begin by simplifying the expression before we differentiate it. ) Thus,
[pic][pic]
( The outer layer is ``the negative four-fifths power'' and the inner layer is [pic]. Differentiate ``the negative four-fifths power'' first, leaving [pic]unchanged. Then differentiate [pic]. )
[pic]
[pic]
(At this point, we will continue to simplify the expression, leaving the final answer with no negative exponents.)
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic].
SOLUTION 6 : Differentiate [pic].
( 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,
[pic]
[pic]
[pic].
SOLUTION 7 : Differentiate [pic].
( The outer layer is ``the exponential function'' and the inner layer is [pic]. Recall that [pic]. Differentiate ``the exponential function'' first, leaving [pic]unchanged. Then differentiate [pic]. ) Thus,
[pic]
[pic]
[pic].
SOLUTION 8 : Differentiate [pic].
( The outer layer is ``2 raised to a power'' and the inner layer is [pic]. Recall that [pic]. Differentiate ``2 raised to a power'' first, leaving [pic]unchanged. Then differentiate [pic]. ) Thus,
[pic]
[pic]
[pic].
SOLUTION 9 : Differentiate [pic].
( Since 3 is a MULTIPLIED CONSTANT, we will first use the rule [pic], 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,
[pic]
( Now the outer layer is ``the tangent function'' and the inner layer is [pic]. Differentiate ``the tangent function'' first, leaving [pic]unchanged. Then differentiate [pic]. )
[pic]
[pic]
[pic]
[pic]
[pic].
SOLUTION 10 : Differentiate [pic].
( The outer layer is ``the natural logarithm (base e) function'' and the inner layer is ( 17-x ) . Recall that [pic]. Differentiate ``the natural logarithm function'' first, leaving ( 17-x ) unchanged. Then differentiate ( 17-x ). ) Thus,
[pic]
[pic]
[pic].
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