### SOLUTIONS TO DIFFFERENTIATION OF FUNCTIONS USING THE PRODUCT RULE

SOLUTION 13 : Consider the function . For what values of x is f'(x) = 0 ? Begin by differentiating the function using the product rule. Then

= 0 .

It follows that

or 4x+3 = 0 .

Thus, the values of x which solve f'(x) = 0 are

or .

SOLUTION 14 : Consider the function . For what values of x is f'(x)=0 ? Begin by differentiating the function using the product rule. Then

= 0 .

It follows that

or -x+1 = 0 .

But can never be zero since an exponential is always positive. Thus, the only values of x which solve f'(x) = 0 are

x = 0 or x = 1 .

SOLUTION 15 : Consider the function . For what values of x is f'(x)=0 ? Begin by differentiating the function using the product rule. Then

= 0 .

It follows that

x = 0 or .

But x = 0 is not in the domain of function f since is not defined. Thus, the only possibility is

iff

iff

iff

iff

.

SOLUTION 16 : Prove that

.

Group functions f and g and apply the ordinary product rule twice. Then

= f'(x) g(x) h(x) + f(x) g'(x) h(x) + f(x) g(x) h'(x) .

Here is an easy way to remember the triple product rule. Each time differentiate a different function in the product. Then add the three new products together.

SOLUTION 17 : Differentiate . Differentiate y using the triple product rule. Then

.

SOLUTION 18 : Consider the function . For what values of x is f'(x)=0 ? Begin by differentiating f using the triple product rule. Then

(Now factor out the common terms of , and .)

.

It follows that

or .

But is never zero since exponentials are always positive. Thus, the values of x which solve f'(x)=0 are

or (using the quadratic formula) .

SOLUTION 19 : 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

.

The slope of the line tangent to the graph at is

.

Thus, an equation of the tangent line is

or y=x .

SOLUTION 20 : Find an equation of the line perpendicular to the graph of at . If then so that the tangent line passes through the point . The slope of the line perpendicular to the graph of f will be perpendicular to the line tangent to the graph of f . Thus, we need to first compute the derivative of f . Then

.

Thus, the slope of the line tangent to the graph at is

.

Thus, the slope of the line perpendicular to the graph at is

and an equation of this line is

.

SOLUTION 21 : Find all points (x, y) on the graph of with tangent lines parallel to the line y + x = 12 . The slope of the given line y = 12-x is y'=-1 . Thus, we need to find all values x in the domain of f which satisfy f'(x)=-1 . The derivative of f is

.

If then . It follows that

iff

iff

iff

iff

x(9x + 8) = 0 .

Thus,

x = 0 or 9x + 8 = 0

so that

x = 0 or .

But x=0 cannot be a solution since . Therefore, the only solution is the point

and

.