### SOLUTIONS TO DIFFERENTIATION OF FUNCTIONS USING THE QUOTIENT RULE

SOLUTION 12 : Differentiate .

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 .)

.

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).)

.

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

(Now apply the quotient rule.)

.

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

(Recall that and apply the quotient rule.)

(Factor x and from the numerator.)

.

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 = -5x - 4 .

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

.

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 2x and in the numerator.)

(Recall that and .)

.

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

Thus,

2x (1-x) = 0 ,

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

so that

2x = 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 .

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

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

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