. Apply the product rule.
Then
(Factor an x from each term.)
.
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SOLUTION 2 : Differentiate 
. Apply the quotient rule. Then
.
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SOLUTION 3 : Differentiate 
arc
arc
.
Apply the product rule. Then
arc
arc
arc
arc
arc
arc
= ( arc
arc
.
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SOLUTION 4 : Let 
arc
. Solve f'(x) = 0 for x . Begin by differentiating f . Then
(Get a common denominator and subtract fractions.)
.
(It is a fact that if 
, then A = 0 .) Thus,
2(x - 2)(x+2) = 0 .
(It is a fact that if AB = 0 , then A = 0 or B=0 .) It follows that
x-2 = 0 or x+2 = 0 ,
that is, the only solutions to f'(x) = 0 are
x = 2 or x = -2 .
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SOLUTION 5 : Let 
. Show that f'(x) = 0 . Conclude that
. Begin by differentiating f . Then
.
If f'(x) = 0 for all admissable values of x , then f must be a constant function, i.e.,
for all admissable values of x ,
i.e.,
for all admissable values of x .
In particular, if x = 0 , then
i.e.,
.
Thus, 
and for all admissable values of x .
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SOLUTION 6 : Evaluate 
. It may not be obvious, but this problem can be viewed as a derivative problem. Recall that
(Since h approaches 0 from either side of 0, h can be either a positve or a negative number. In addition,
is equivalent to 
. This explains the following
equivalent variations in the limit definition of the derivative.)
.
If 
, then 
, and letting
, it follows that
.
The following problems require use of the chain rule.
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SOLUTION 7 : Differentiate 
. Use the product rule first. Then
(Apply the chain rule in the first summand.)
(Factor out 
. Then get a common denominator and add.)
.
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SOLUTION 8 : Differentiate 
. Apply the chain rule twice. Then
(Recall that 
.)
.
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SOLUTION 9 : Differentiate 
. Apply the chain rule twice. Then
(Recall that 
.)
.
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SOLUTION 10 : Determine the equation of the line tangent to the graph of 
at x = e . If x = e , then 
, so that the line passes through the point 
. The slope of the tangent line follows from the derivative (Apply the chain rule.)
.
The slope of the line tangent to the graph at x = e is
.
Thus, an equation of the tangent line is
.
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