## Summary of Chapter 4 inApplied CalculusChapter 11 inFinite Mathematics & Applied CalculusTopic: Techniques of Differentiation

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Product & Quotient Rules | Calculation Thought Experiment | Chain Rule | Derivatives of Logarithmic and Exponential Functions | Derivatives of Trigonometric Functions | Implicit Functions & Implicit Differentiation

Product Rule
 ddx [f(x)g(x)] = f'(x) g(x) + f(x)g'(x)

Product Rule In Words:

The derivative of a product is the derivative of the first times the second, plus the first times the derivative of the second.

Quotient Rule

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

Quotient Rule In Words:

The derivative of a quotient is the derivative of the top times the bottom, minus the top times the derivative of the bottom, all over the bottom squared..
Examples

Product Rule

 ddx [x2(3x- 1)] = 2x(3x- 1) + x2(3)

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

Quotient Rule

 ddx x3x2+1 = 3x2(x2+1) - x3.2x(x2+1)2

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

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Combining Rules for Differentiation: Calculation Thought Experiment

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...

Examples

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):
 ddx [x2 + (4x- 1)(x+2)] = ddx [x2] + ddx [(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:

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

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Chain Rule

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

 ddx [f(u)] = f'(u) dudx

Chain Rule In Words:

The derivative of f(quantity) is the derivative of f, evaluated at that quantity, times the derivative of the quantity.

For instance, if f(u) = u0.5, then
 ddx [u0.5] = 0.5 u- 0.5 dudx

The derivative of a quantity raised to the 0.5 is 0.5 times the quantity raised to the -0.5, times the derivative of the quantity

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)
 ddx xn = nx n- 1
 ddx un = nun- 1 dudx
 ddx 4x- 1/2 = - 2x- 3/2
 ddx 4u- 1/2 = - 2u- 3/2 dudx
 ddx sin x = cos x
 ddx sin u = cos u dudx
Examples

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

1.
d

dx
(1+x2)3 =
 3(1+x2)2 ddx (1+x2)
=
 3(1+x2)2,2x
=
 6x(1+x2)2
2.
d

dx
2

(x+x2)3
=
 ddx (x+x2)- 3
=
 - 3(x+x2)- 4 ddx (x+x2)
=
 - 3(x+x2)- 4 (1+2x)
=
 1+2x(x+x2)4
3.
d

dx
e(x+x2) =
 e(x+x2) ddx (x+x2)
=
 e(x+x2)(1+2x)
4.
d

dx
(x- 2/x)- 1 =
 Use proper graphing calculator format

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Derivatives of Logarithmic and Exponential Functions

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)
 ddx ln x   = 1x
 ddx ln u  = 1u dudx
 ddx logax = 1x ln a
 ddx logau = 1u ln a dudx
 ddx ex = ex
 ddx eu = eu dudx
 ddx ax = axln a
 ddx au = auln a dudx
Examples

1.
d

dx
2 ln x =
 2 1x = 2x
2.
d

dx
ln(2x) =
 12x ddx (2x)
=
 2 12x = 1x
3.
d

dx
log3(2x+1) =
 1(2x+1)ln 3 ddx (2x+1)
=
 2 1(2x+1)ln 3 = 2(2x+1)ln 3
4.
d

dx
ex2+1 =
 ex2+1 ddx (x2+1)
=
 2x ex2+1
5.
d

dx
2(x3+x) =
 Use proper graphing calculator format

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Derivatives of Trigonometric 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)
 ddx sin x = cos x
 ddx sin u = cos u dudx
 ddx cos x = - x
 ddx cos u = - sin u dudx
 ddx tan x = sec2 x
 ddx tan u = sec2u dudx
 ddx cotan x = - cosec2 x
 ddx cotan u = - cosec2u dudx
 ddx sec x = sec x tan x
 ddx sec u = sec u tan u dudx
 ddx cosec x = - cosec x cotan x
 ddx cosec u = - cosec u cotan u dudx
Example

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) ddx (2x2+1)
=
 sin(2x2+1).4x = 4x sin(2x2+1)
3.
d

dx
sec(x3) =
 sec(x3) tan(x3) ddx (x3)
=
 sec(x3) tan(x3) . 3x2
=
 3x2 sec(x3) tan(x3)
5.
d

dx
x cos(x2) =
 Use proper graphing calculator format

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Implicit Functions and Implicit Differentiation

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, and then substitute the values of x and y in the expression for dy/dx.

Logarithmic differentiation 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.

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Last Updated: April 2006