Then

(Apply the product rule in the second part of the numerator. Recall that .)

(Begin to simplify the final answer by getting a common denominator in the numerator.)

(Simplify the numerator and invert and multiply by the reciprocal of the denominator.)

(Recall that .)

.

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* SOLUTION 13 :* Differentiate . First apply the chain rule. Then

(Now apply the quotient rule and recall that . Please not that there is a TYPO in the remaining steps of this solution. Every subsequent term (3x+1) should be (3x+2).)

(Recall that and .)

(Recall that .)

(Please note that there is a TYPO in the final answer. The term (3x+1) should be (3x+2).)

.

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* SOLUTION 14 :* Differentiate .
Recall that and apply the chain rule. Then

(Now apply the quotient rule.)

.

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* SOLUTION 15 :* Differentiate . Apply the product rule first. Then

(Recall that and apply the quotient rule.)

(Begin to simplify your answer by finding a common denominator.)

(Factor *x* and from the numerator.)

.

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* SOLUTION 16 :* Find an equation of the line tangent to the graph of at *x*=-1 . If *x*= -1 then
so that the tangent line passes through the point
(-1, 1 ) . The slope of the tangent line follows from the derivative of *y* . Then

.

The slope of the line tangent to the graph at *x* = -1 is

.

Thus, an equation of the tangent line is

*y* - 1 = -5 (*x* - (-1) ) or *y* = -5*x* - 4 .

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* SOLUTION 17 :* Find an equation of the line tangent to the graph of at . If then
so that the tangent line passes through the point
. The slope of the tangent line follows from the derivative of *y* . Then

.

There is no need to simpliy this derivative. Just let in the derivative equation. Thus, the slope of the line tangent to the graph at is

*m* = *y*'

.

Thus, an equation of the tangent line is

.

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* SOLUTION 18 :* Consider the function .
Solve *f*'(*x*) = 0 for *x* . Solve *f*''(*x*) = 0 for *x* . The first derivative is

(Factor out the common terms of 2*x* and in the numerator.)

(Recall that and .)

.

(It is a fact that if then *A* = 0 .)

Thus,

2*x* (1-*x*) = 0 ,

(It is a fact that if *A B* = 0 , then *A*=0 or *B* = 0 .)

so that

2*x* = 0 or 1-*x* = 0 ,

i.e.,

*x* = 0 or *x* = 1

are the only solutions to *f*'(*x*) = 0 .

The second derivative is

(Factor from the numerator.)

.

It follows that

,

so that using the quadratic formula we get that

or

are the only solutions to *f*''(*x*) = 0 .

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* SOLUTION 19 :* Find all points (*x*, *y*) on the graph of
where tangent lines are perpendicular to the line 8*x*+2*y* = 1 . Since

8*x*+2*y* = 1 iff 2*y* = 1 - 8*x* iff ,

the slope of this given line is -4. Thus, a tangent line to the graph of *f* which is perpendicular to
8*x*+2*y* = 1 will have slope

.

The slope of a tangent line is also given by the derivative

.

Thus, equating slopes, it follows that

iff

iff

(Recall that .)

iff

| 2-*x* | = 2

iff

iff

*x* = 0 or *x* = 4 .

It follows that there are two distinct tangent lines which are perpendicular to the line 8*x*+2*y* = 1 .
If *x* = 0 , then

,

and if *x* = 4 , then

,

so that the two distinct points of tangency are

( 0, -1/ 2 ) and ( 4, -3/ 2 ) .

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Wed Jul 30 20:10:53 PDT 1997