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Derivatives of Trigonometric Functions for AP Calculus

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By — McGraw-Hill Professional
Updated on Oct 24, 2011

Practice problems for these concepts can be found at: Differentiation Practice Problems for AP Calculus

Summary of Derivatives of Trigonometric Functions

Note that the derivatives of cosine, cotangent, and cosecant all have a negative sign.

Example 1

If y =6x2 + 3 sec x, find .

=12x + 3 sec x tan x.

Example 2

Find f '(x) if f(x) = cot(4x – 6).

Using the chain rule, let u = 4x – 6. Then f '(x ) = [–csc2(4x – 6)] [4] = –4 csc2(4x – 6).

Or using your calculator, enter d(1/ tan(4x – 6), x ) and obtain which is an equivalent form.

Example 3

Find f '(x) if f(x ) = 8 sin(x2).

Using the chain rule, let u = x2. Then f '(x) = [8 cos(x2)] [2x] = 16x cos(x2).

Example 4

If y = sin x cos(2x), find .

Using the product rule, let u = sin x and v = cos(2x).

Then = cos x cos(2x) + [–sin(2x)](2)(sin x) = cos x cos(2x) – 2 sin x sin(2x).

Example 5

If y = sin[cos(2x)], find .

Using the chain rule, let u = cos(2x). Then

= cos[cos(2x)] [cos(2x)].

To evaluate [cos(2x)], use the chain rule again by making another u-substitution, this time for 2x. Thus, [cos(2x)] = [–sin(2x)]2 = – 2 sin(2x). Therefore, cos[cos(2x)](–2 sin(2x))= – 2 sin(2x) cos[cos(2x)].

Example 6

Find f '(x) if f(x) = 5x csc x.

Using the product rule, let u =5x and v = csc x. Then f '(x) = 5 csc x + (–csc x cot x) (5x) = 5 csc x – 5x(csc x)(cot x).

Example 7

If y = , find

Rewrite y = as y = (sin x)1/2. Using the chain rule, let u = sin x. Thus,

Example 8

If y =

Using the quotient rule, let u = tan x and v = (1+ tan x ). Then,

Note: For all of the above exercises, you can find the derivatives by using a calculator, provided that you are permitted to do so.

Practice problems for these concepts can be found at: Differentiation Practice Problems for AP Calculus

 

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