Power Rule
If f(x) = xⁿ, then f'(x) = nx^n-1. This is true for any real n. In differential notation, this reads

d dx xn

=

nxn - 1

Sum and Constant Multiple Rules
If f'(x) and g'(x) exist, and c is a constant, then

(A) [f(x)   ±   g(x)]' = f'(x)   ±   g'(x)

(B) [cf(x)]' = cf'(x) .

(A)

d dx

[f(x) ± g(x)] =

d dx

[f(x)] ±

d dx

[g(x)]

(B)

d dx

[cf(x)] = c

d dx

[f(x)]

In words:

The derivative of a sum is the sum of the derivatives, and the derivative of a difference is the difference of the derivatives.

The derivative of c times a function is c times the derivative of the function.

d dx x3.2

=

3.2x2.2

d dx 4x - 1

=

- 4x - 2

d dx 3x1.1 + 4x - 2

=

3.3x0.1 - 8x - 3

d dx 1 x	=	d dx x - 1
	=	- x - 2
	=	- 1 x2

Want some practice? Try the interactive tutorial or try some exercises.

If Q(x) represents any quantity such as cost, revenue, profit or loss on the sale of x items, then Q'(x) is called the marginal quantity. Thus, for instance, the marginal cost measures the increase in total cost per item. This is effectively the cost of each additional item.

The marginal cost is distinct from the average cost, which measures the average of the total cost of the first x items. Average cost is given by

C(x)	=	C(x) x
	=	Total Cost Number of Items

Suppose the cost of (the first) x items is given by

C(x) = 4x^0.2 - 0.1x pounds Sterling.

C'(x) = 0.8x^-0.8 - 0.1 pounds Sterling per unit.

The average cost of the first three units is

C(3)	=	C(3) 3
	≈	4.6829 3
	≈	1.56 pound Sterling/item

d dx

[f(x)g(x)]

=

f'(x)g(x) + f(x)g'(x)

Product Rule In Words:

Quotient Rule

d dx

f(x) g(x)

=

f'(x)g(x) - f(x)g'(x) g(x)2

Quotient Rule In Words:

Product Rule

d dx

[x2(3x- 1)]

=

2x(3x- 1) + x2(3)

(The derivatives of f and g are shown in color.)

Quotient Rule

d dx

x3 x2+1

=

3x2(x2+1) - x3.2x (x2+1)2

Of course, you should simplify the answers and not leave them like that!

Press here for an on-line tutorial on the product & quotient rules.

The calculation thought experiment is a technique to determine whether to treat an algebraic expression as a product, quotient, sum, or difference. Given such an expression, consider the steps you would use in computing its value. If the last operation is multiplication, treat the expression as a product; if the last operation is division, treat the expression as a quotient, and so on.

Using the Calculation Thought Experiment (CTE) to Differentiate a Function
If the CTE says, for instance, that the expression is a sum of two smaller expressions, then apply the rule for sums as a first step. This will leave you having to differentiate simpler expressions, and you can use the CTE on these, and so on...

1. (3x²- 4)(2x+1) can be computed by first calculating the expressions in parentheses and then multiplying. Since the last step is multiplication, we can treat the expression as a product.

2. (2x- 1)/x can be computed by first calculating the numerator and denominator, and then dividing one by the other. Since the last step is division, we can treat the expression as a quotient.

3. x² + (4x- 1)(x+2) can be computed by first calculating x², then calculating the product (4x- 1)(x+2), and finally adding the two answers. Thus, we can treat the expression as a sum.

4. (3x²- 1)⁵ can be computed by first calculating the expression in parentheses, and then raising the answer to the fifth power. Thus, we can treat the expression as a power.

Using the CTE
Let us differentiate x² + (4x- 1)(x+2). Since we saw above that this is the sum of x² and (4x- 1)(x+2), the first step is to use the rule for differentiating a sum (discussed in the preceding topic summary):

d dx

[x2 + (4x- 1)(x+2)]

=

d dx

[x2]

+

d dx

[(4x- 1)(x+2)]

Now we are left with two simpler functions to differentiate: x², which is a power, so we use the power rule, and (4x- 1)(x+2), which is a product, so that we use the product rule on this:

d dx [x2] + d dx [(4x- 1)(x+2)]

= 2x + 4(x+2) + (4x- 1)(1)

If f is a differentiable function of u and u is a differentiable function of x, then the composite f(u) is a differentiable function of x, and

d dx

[f(u)]

=

f'(u)

du dx

Chain Rule In Words:

d dx

[u0.5]

=

0.5u- 0.5

du dx

Generalized Differentiation Rules
As the above Example illustrates, for every function of u whose derivative we know we now get a "generalized" differentiation rule:

Original Rule	Generalized Rule (Chain Rule)
d dx xn = nx n- 1	d dx un = nun- 1 du dx
d dx 4x- 1/2 = - 2x- 3/2	d dx 4u- 1/2 = - 2u- 3/2 du dx
d dx sin x = cos x	d dx sin u = cos u du dx

(First look over the generlized rules on the left.)

1.	d dx	(1+x2)3	=	3(1+x2)2 d dx (1+x2)
			=	3(1+x2)2,2x
			=	6x(1+x2)2
2.	d dx	2 (x+x2)3	=	d dx (x+x2)- 3
			=	- 3(x+x2)- 4 d dx (x+x2)
			=	- 3(x+x2)- 4 (1+2x)
			=	−3(1+2x) (x+x2)4
3.	d dx	e(x+x2)	=	e(x+x2) d dx (x+x2)
			=	e(x+x2)(1+2x)
4.	d dx	(x- 2/x)- 1	=	Use proper graphing calculator format

Press here for an on-line tutorial on the chain rule.

The following table summarizes the derivatives of logarithmic and exponential functions, as well as their chain rule counterparts (that is, the logarithmic and exponential functions of a function).

Original Rule	Generalized Rule (Chain Rule)
d dx ln x   = 1 x	d dx ln u  = 1 u du dx
d dx logax = 1 x ln a	d dx logau = 1 u ln a du dx
d dx ex = ex	d dx eu = eu du dx
d dx ax = axln a	d dx au = auln a du dx

1.	d dx	2 ln x	=	2. 1 x = 2 x
2.	d dx	ln(2x)	=	1 2x d dx (2x)
			=	2. 1 2x = 1 x
3.	d dx	log3(2x+1)	=	1 (2x+1)ln 3 d dx (2x+1)
			=	2. 1 (2x+1)ln 3 = 2 (2x+1)ln 3
4.	d dx	ex2+1	=	ex2+1 d dx (x2+1)
			=	2x ex2+1
5.	d dx	2(x3+x)	=	Use proper graphing calculator format

Press here for an on-line tutorial on derivatives of logarithmic and exponential functions.

The following table summarizes the derivatives of the six trigonometric functions, as well as their chain rule counterparts (that is, the sine, cosine, etc. of a function).

Original Rule	Generalized Rule (Chain Rule)
d dx sin x = cos x	d dx sin u = cos u du dx
d dx cos x = - x	d dx cos u = - sin u du dx
d dx tan x = sec2 x	d dx tan u = sec2u du dx
d dx cotan x = - cosec2 x	d dx cotan u = - cosec2u du dx
d dx sec x = sec x tan x	d dx sec u = sec u tan u du dx
d dx cosec x = - cosec x cotan x	d dx cosec u = - cosec u cotan u du dx

1.	d dx	x sin x	=	1.sin x + x cos x   Product rule
			=	sin x + x cos x
2.	d dx	cos(2x2+1)	=	sin(2x2+1) d dx (2x2+1)
			=	sin(2x2+1).4x = 4x sin(2x2+1)
3.	d dx	sec(x3)	=	sec(x3) tan(x3) d dx (x3)
			=	sec(x3) tan(x3) . 3x2
			=	3x2 sec(x3) tan(x3)
5.	d dx	x cos(x2)	=	Use proper graphing calculator format

Press here for on-line text on derivatives of trigonometric functions.

Given an equation in x and y, we can think of y as an implicit function of x. We can find dy/dx without first solving for y as follows:

• First, take the derivative with respect to x of both sides of the equation (treating y as "a quantity" in the chain rule).
• Next, solve for dy/dx. This may give dy/dx in terms of both x and y.
• To evaluate dy/dx at a specific value of x (or y), first substitute the given value in the original equation relating x and y to obtain a value for the other variable if necessary, and then substitute the values of x and y in the expression for dy/dx.

Logarithmic differentiation
This is the technique of first taking the (natural) logarithm of both sides of an equation and then finding dy/dx using implicit differentiation. Logarithmic differentiation is a useful alternative to the product and quotient rules when finding derivatives of particularly complicated expressions.

Suppose we want to find dy/dx given that

xy+ y3x -ey = 4

d dx [xy + y3x -ey] = d dx [4]

y + x dy dx + 3y2 dy dx x + y3 - ey dy dx = 0

x

dy dx

+ 3y2

dy dx

x - ey

dy dx

=

-y - y3

dy dx   x + 3xy2 - ey   = -y - y3

dy dx = -y - y3 x + 3xy2 - ey