= 0 .

It follows that

or 4*x*+3 = 0 .

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

or .

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* 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 .

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* 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

.

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* 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.

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

.

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* 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) .

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* 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* .

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* 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

.

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* 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*(9*x* + 8) = 0 .

Thus,

*x* = 0 or 9*x* + 8 = 0

so that

*x* = 0 or .

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

and

.

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Fri May 9 12:13:55 PDT 1997