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* =6*x*^{2} + 3 sec *x*, find .

=12*x* + 3 sec *x* tan *x*.

**Example 2**

Find *f* '(*x*) if *f*(*x*) = cot(4*x* – 6).

Using the chain rule, let *u* = 4*x* – 6. Then *f* '(*x* ) = [–csc^{2}(4*x* – 6)] [4] = –4 csc^{2}(4*x* – 6).

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

**Example 3**

Find *f* '(x) if *f*(*x* ) = 8 sin(*x*^{2}).

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

**Example 4**

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

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

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

**Example 5**

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

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

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

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

**Example 6**

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

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