● F

Example: Find derivative if

Later, though, we will meet functions, such as y = x2 sinx, for which the product rule is the only possible method.

● F

example:

Product and QuotientDifferentiation Rules

The Quotient Rule can be used to extend the Power Rule to the case where the exponent is a negative integer.

1( ) n ndx nx

dx

Example:

● F n positive integer

▪ If n = 0, then x0 = 1, which we know has a derivative of 0. Thus, the Power Rule holds for any integer n. What if the exponent is a fraction?

▪ In fact, it can be shown by using Chain Rule (obviously proof later) that it also holds for any real number n.

If is any real number, then

1( )n ndx nx

dx

● F

The derivatives of the remaining trigonometric functions — csc, sec, and cot — can also be found easily using the Quotient Rule.

All together:

Example:

Differentiate For what values of x does the graph of

f have a horizontal tangent?

Since sec x is never 0, we see that f’(x) = 0 when tan x = 1.

This occurs when x = nπ + π/4, where n is an integer

Example:

Find

Example:

Calculate:

If is continuous on and is any number between and , then there is at least one number such that .

example:

Prove that function has a root/zero between 2 and 2.5.

is continuous on ], and , so must have a zero between 2 and 2.5.

Intermediate Value Theorem for Continuous Functions

As example on our graph:As example on our graph:

Between and , must take on every value between ½ and 3.

If and are any two points in an interval on which is differentiable, then takes on every value between and .

You can find it in this form too:

Let be differentiable on and suppose that k is a number between and . Then there exists a point such that .

is continuous function on

Intermediate Value Theorem for Derivatives

𝑑𝑑𝑥

𝑐=0

𝑑𝑑𝑥

𝑥𝑛=𝑛𝑥𝑛−1

𝑑𝑑𝑥sin 𝑥=cos𝑥

𝑑𝑑𝑥cos𝑥=−sin 𝑥

𝑑𝑑𝑥tan 𝑥=𝑠𝑒𝑐2𝑥

𝑑𝑑𝑥cot𝑥=−𝑐𝑠𝑐2𝑥

𝑑𝑑𝑥csc𝑥=−csc 𝑥cot 𝑥

𝑑𝑑𝑥sec𝑥=sec 𝑥 tan𝑥

(𝑐𝑓 )′=𝑐𝑓 ′

( 𝑓 +𝑔 )′= 𝑓 ′+𝑔 ′

( 𝑓 −𝑔 )′= 𝑓 ′−𝑔 ′

( 𝑓𝑔 )′= 𝑓 ′𝑔+ 𝑓𝑔 ′

( 𝑓𝑔 )′

= 𝑓 ′𝑔− 𝑓𝑔 ′  𝑔2

Higher Order Derivatives:

Just as one can obtain a velocity function by differentiating a position function, one can obtain an acceleration function by differentiating a velocity function. Alternatively, one can think about obtaining an acceleration function by differentiating a position function twice.

The notation is called the second derivative of and we can read it as

Higher Order Derivatives:

Recognizing a given limit as a derivative (!!!!!!)

• From a general definition of the derivative: FORM A

𝑓 ′ (𝑥 )=𝑙𝑖𝑚h→ 0

𝑓 (𝑥+h)− 𝑓 (𝑥)h

limh→0

3√8+h−2h

= 𝑑𝑑𝑥

3√𝑥|𝑥=8

=13𝑥− 2/3|

𝑥=8

= 112

Example:

Example:

𝑙𝑖𝑚 𝑥→𝑐

𝑓 (𝑥 )− 𝑓 (𝑐 )𝑥−𝑐

= 𝑓 ′ (𝑐 )(𝑙𝑖𝑚 h→0

𝑓 (𝑐+h )− 𝑓 (𝑐)(𝑐+h)−𝑐

= 𝑓 ′ (𝑐 ))• From the definition of the derivative at point c: FORM B

¿−sin 𝑥|𝑥=𝜋

3

=− √32