### SOLUTIONS TO DIFFERENTIATION OF TRIGONOMETRIC FUNCTIONS

SOLUTION 1 : Differentiate .

(Recall that . The product rule is NOT necessary here.)

Then

.

SOLUTION 2 : Differentiate . Apply the product rule.

Then

.

SOLUTION 3 : Differentiate . Apply the quotient rule.

Then

(Recall the well-known trigonometry identity .)

.

SOLUTION 4 : Differentiate . Apply the product rule.

Then

.

SOLUTION 5 : Differentiate . To avoid using the chain rule, first rewrite the problem as

.

Now apply the product rule. Then

.

SOLUTION 6 : Differentiate . To avoid using the chain rule, recall the trigonometry identity , and first rewrite the problem as

.

Now apply the product rule twice. Then

(This is an acceptable answer. However, an alternative answer can be gotten by using the trigonometry identity .)

.

SOLUTION 7 : Differentiate . Rewrite g as a triple product and apply the triple product rule. Then

so that the derivative is

.

SOLUTION 8 : Evaluate . It may not be obvious, but this problem can be viewed as a differentiation problem. Recall that

.

If , then , and letting it follows that

.

SOLUTION 9 : Differentiate . Apply the chain rule to both functions. (If necessary, review the section on the chain rule .) Then

(Recall that .)

.

SOLUTION 10 : Differentiate . This is NOT a product of functions. It's a composition of functions. Apply the chain rule. Then

.

SOLUTION 11 : Differentiate . Apply the quotient rule first, followed by the chain rule. Then

.

SOLUTION 12 : Differentiate . Apply the product rule first, followed by the chain rule. Then

.

SOLUTION 13 : Differentiate . Apply the chain rule four times ! Then

.

SOLUTION 14 : Differentiate . Apply the quotient rule first. Then

(Apply the product rule in the first part of the numerator.)

.

SOLUTION 15 : 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, 0 ) . The slope of the tangent line follows from the derivative

.

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

= -2 .

Thus, an equation of the tangent line is

y - 0 = -2 (x - (-1) ) or y = -2x - 2 .

SOLUTION 16 : 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 tangent line follows from the derivative of y . Then

.

The slope of the line tangent to the graph at is

.

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

m = - 2 ,

so that an equation of the line perpendicular to the graph at is

or .

SOLUTION 17 : Assume that . Solve f'(x) = 0 for x in the interval . Use the chain rule to find the derivative of f . Then

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

so that

or .

If , then the only solutions x in are

or .

If , then the only solutions x in are

or .

Thus, the only solutions to f'(x) = 0 in the interval are

or .

SOLUTION 18 : Use any method to verify that .

Then

(Apply the quotient rule.)

(Recall the well-known trigonometry identity .)

(Recall that .)

.