Page 1

Can you see the pattern?

And indeed, (Bring the exponent to the front as a coefficient, and reduce the exponent by 1, and use that as the new exponent.)

This is our very first formula in derivatives without using the limit definition. It works for any base (as long as the base is the variable and nothing else) and any exponent (as long as the exponent is a real number). Technically, in mathematics, we use the letter n to denote a positive integer, but since this formula applies for any real number exponent, so it’s better to rewrite the formula as:

Use the above formula to find the derivative of the following functions:

For negative exponents: For rational exponents: For irrational exponents:

2( ) 2d

x xdx

( ) 1d

xdx

3 2( ) 3d

x xdx

5( )d

xdx

4( )d

xdx

6( )d

xdx

( )ndx

dx

1( ) (where is a real number)r rdx r x r

dx

2

10

2007

1.

2.

13.

14.

dx

dxd

xdxd

dx x

d

dx x

Finally a Formula for Derivative and Its Application

However, the formula is NOT applicable for the following examples:

1. 2. 3.

3/ 2

2/3

3 4

1.

2.

3.

4.

dx

dxd

xdxd

xdxd

xdx

2

2 3 4 5

1.

2.

3.

4.

e

e

dx

dxd

xdxd

xdxd

xdx

1( )r rdx r x

dx

3[(2 1) ]d

xdx

5 6( )xdx

dx (4 )xd

dx

Page 2

Derivative of a Constant FunctionIf f(x) = 7, what is f (x)? And in general, If f(x) = k where k is a constant, what is f (x)?

Limits Laws Derivatives Formulas

1.

2.

3.

[ ( )] [ ( )]d d

k f x k f xdx dx

[ ( ) ( )] [ ( )] [ ( )]d d d

f x g x f x g xdx dx dx

lim[ ( )] lim[ ( )]x a x a

k f x k f x

lim[ ( ) ( )] lim[ ( )] lim[ ( )]x a x a x c

f x g x f x g x

lim[ ( ) ( )] lim[ ( )] lim[ ( )]x a x a x a

f x g x f x g x

[ ( ) ( )] [ ( )] [ ( )]d d d

f x g x f x g xdx dx dx

Derivative of a Constant Function and Some Derivative Formulas

Why the formula is NOT applicable for the following examples?

1. 2. 3.

1( )r rdx r x

dx

3[(2 1) ]d

xdx

5 6( )xdx

dx (4 )xd

dx

Some Fundamental Derivative FormulasIn limits, we have laws. In derivatives we have formulas too.

Applications:1. If f(x) = 2x3 – 4x2 + 5x – 6, what is f (x)? 2. Given g(t) = t6 + t–19 + t–2007, find g (t). 3.

197

6 200dx

dx x x

Keep this in mind too:

1

1

1.

2.

3.

4.

5.

r r

r r

dx r x

dxd

k x kr xdxd

kdxd

xdxd

kxdx

Page 3

Limits Laws Derivatives Formulas In Words…

1.The limit/derivative of a constant times a function is the constant times the limit/derivative of the function.

2.The limit/derivative of a sum of two functions is the sum of the limit/derivative of the two functions.

3.The limit/derivative of a difference of two functions is the difference of the limit/derivative of the two functions.

[ ( )] [ ( )]d d

k f x k f xdx dx

[ ( ) ( )] [ ( )] [ ( )]d d d

f x g x f x g xdx dx dx

lim[ ( )] lim[ ( )]x a x a

k f x k f x

lim[ ( ) ( )] lim[ ( )] lim[ ( )]x a x a x c

f x g x f x g x

lim[ ( ) ( )] lim[ ( )] lim[ ( )]x a x a x a

f x g x f x g x

[ ( ) ( )] [ ( )] [ ( )]d d d

f x g x f x g xdx dx dx

Derivative of a Product

In the previous page, we showed you three formulas in derivatives, which really are complementary to three laws in limits.

In limits, we have the limit of a product of two functions is the product of the limit of the two functions:

You might wonder: In derivatives, is the derivative of a product of two functions the product of the derivative of the two functions:

)]([lim)]([lim)]()([lim xgxfxgxfaxaxax

)]([)]([)]()([ xgdx

dxf

dx

dxgxf

dx

d

If it is, then d/dx[x2x3] =

But d/dx[(x2)(x3)] =

If it is, then d/dx[2x3] =

But d/dx[2x3] =

f(x) g(x)

f(x) g(x)

The two examples above show that the derivative of a product ______ the product of the derivatives. That is,

)]([)]([)]()([ xgdx

dxf

dx

dxgxf

dx

d

Page 4The Product Rule

If the derivative of a product is not the product of the derivatives, then what is the derivative of a product of two functions:

?)]()([)]([)]([)]()([ xgxfdx

dxg

dx

dxf

dx

dxgxf

dx

d

It turns out the correct formula should be:

)()()()()()( xgdx

dxfxgxf

dx

dxgxf

dx

d

and this is called the Product Rule.

Keep the second factor function

Differentiate the second factor function

Differentiate the first factor function

Keep the first factor function

What did we just do: times

plus

times

Why it works? Let’s see, with the product rule, we have:

4532444223323232 5][][and53232][][][ xxdx

dxx

dx

dxxxxxxxx

dx

dxxx

dx

dxx

dx

d Check!

23333 6]2[and][2]2[]2[ xxdx

dx

dx

dx

dx

dx

dx

d

)]54)(32[(and)]54)(32[( xxdx

dxx

dx

d

You might say: Why do we need the product rule if it’s easier without it? Answer: Because we do need it sometimes, maybe a lot of the times, if not always.

)]765)(432[(and)]765)(432[( 2222 xxxxdx

dxxxx

dx

dIf you manage to get the derivative of the one above without using the product rule, let’s see how you handle this one:

Page 5If There Is a Product Rule Then There Is a …

If , then what is ? Ans: Definitely NOT .

The correct formula is: and this is called the __________________.

Let’s see how it works on some examples and compare it with the WRONG quotient rule too.

The RIGHT Quotient Rule vs. The WRONG Quotient Rule How Do We Verify Which One Is Right?

)()()()()()( xgdx

dxfxgxf

dx

dxgxf

dx

d

)(

)(

xg

xf

dx

d

)]([

)]([

)(

)(

xg

xf

xg

xf

dx

d

dxddxd

2)]([

)]([)()()]([

)(

)(

xg

xgxfxgxf

xg

xf

dx

d dxd

dxd

x

xx

dx

d 23 2

x

xx

dx

d 23 2

x

xx

dx

d 23 2

4

32 2 xx

dx

d

4

32 2 xx

dx

d

4

32 2 xx

dx

d

34

5

xdx

d

34

5

xdx

d

34

5

xdx

d

Page 6Product Rule and Quotient Rule—In Depth

Sometimes the function might not be in terms of x, the variable we usually use and differentiate with respect to. For example, the function could be in terms of t, in that case, we would have written the product rule as:

Regardless we are using d/dx when we differentiate with respect to x or d/dt when we differentiate with respect to t, we can always use the prime () notation for derivative. Hence, a shorthand for the product rule (by eliminating the x’s and t’s) is:

(fg) = f g + fg

and a shorthand for the quotient rule is: 2

f f g f g

g g

( ) ( ) ( ) ( ) ( ) ( )d d d

f t g t f t g t f t g tdt dt dt

Déjà Vu?Example 1: Given f(x) = (2x + 3)(x2 – 4), find f (x).

Example 2: Given f(x) = , find f (x).2 2 1

2 3

x x

x

Different Versions of Product Rule and Quotient Rule:In some textbooks (including ours), the product rule can be written differently as:

(fg) = f g + gf (fg) = gf + gf (fg) = gf + f

g

(fg) = f g + gf (fg) = gf + gf (fg) = gf + f

g

It’s because addition and multiplication are both commutative, i.e., a + b = b + a and ab = ba, hence we have these different versions.

For quotient rule, you might see these versions in different textbooks:

Our textbook

2

f g f f g

g g

2

f f g g f

g g

2

f g f g f

g g

Our textbook

Page 7Product Rule and Quotient Rule—To Use or Not To Use

My version of the product rule and quotient rule vs. The textbook’s version of the product rule and quotient rule

(fg) = f g + fg (fg) = gf + gf 2

f f g f g

g g

2

f g f f g

g g

If you memorize my version of the product rule, for the quotient rule, all you need to do is to change the “+” sign the “–” sign, and put the whole thing over g2.

If you memorize the textbook version of the product rule, it doesn’t help you to memorize its version of the quotient rule. That means you need to memorize the quotient rule too, which is obviously more complicated.

When to use the product rule and quotient rule and when not to use them?

Product Rule—Examples Use it / Don’t Use it / Doesn’t Matter Quotient Rule—Examples U / DU / DNM

1. f(x) = 2x3

2. f(x) = ¾(x2 + 5x – 6)

3. f(x) = (3x – 1)(2x + 4)

4. f(x) = (3x2 – 2x + 1)(4x2 + 5x – 4)

Change “+” to “–” What is the relationship? Don’t see any.

42

153)(.4

2

153)(.3

4

3)(.2

4

3)(.1

2

2

2

2

x

xxxf

x

xxxf

xxf

xxf