Properties and Formulas of Derivatives

Functions that have Derivatives: We will start off this lesson by learning the relationship between continuity and differentiability.

Definition - Differentiability and Continuity

We are able to tell if a function is continuous at a point if it is differentiable at that point. However, if a function is continuous at a point, then it is not necessarily differentiable at that point.

In short:

  • Differentiable at a point, then continuous at that point
  • Continuous at a point, not necessarily differentiable at that point

Differentiation Theorem

Below are rules we will use a rely on throughout our learning of differentiation

Definition - The Main Differentiation Theorem

For the functions f = f(x) and g = g(x), and f and g are differentiable at x,

Constant Multiple Rule:

For a constant, c,

D(cf(x)) = cD(f(x)) or (cf(x))' = cf'(x)

The derivative of a constant times a function is equal to that constant times the derivative of the function

Sum Rule:

D(f(x) + g(x)) = D(f(x)) + D(g(x)) = D(f)+ D(g)

or

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

The derivative of the sum of two or more functions is equal to the sum of each individual derivative of each of the functions.

Difference Rule:

D(f(x) – g(x)) = D(f(x)) – D(g(x)) = D(f) – D(g)

Or

(f(x) – g(x))' = f'(x) – g'(x)

Similar to the Sum Rule, The derivative of the difference of two or more functions is equal to the difference of each individual derivative of each of the functions.

Product Rule:

D(f(x)g(x)) = f(x)D(g(x)) + g(x)D(f(x)) = fD(g) + gD(f)

Or

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

Quotient Rule:

Dopening bracket
f(x)
g(x)
closing bracket
g(x)D( f(x) ) – f(x)D( g(x))
(g(x))2
gD(f) – fD(g)
g2

or

g(x)f '(x) – f(x)g '(x)
g2

Derivatives of Combinations of Functions

In the example below, we will learn how we can manipulate functions and derivatives using the Differentiation Theorem

Differentiating Combined Functions - Simple Functions

We have two simple functions f(x) and g(x). f(x) = 4 and g(x) = x

D(f(x)) = 0 and D(g(x)) = 1

Calculate the Derivatives of the various combinations and alterations of the functions below.

a) 2g(x)

b) f(x) + g(x)

c) f(x)g(x)

solution to Calculate Derivatives question part a) shown above
solution to Calculate Derivatives question part b) shown above
solution to Calculate Derivatives question part c) shown above

Beginner Differentiation Theorem Questions - Constant Multiple, Sum Rule, Difference Rule, Product Rule, Quotient Rule

Differentiate the following by using the given rule.

a) Use the Constant Multiple Rule to differentiate d over dx derivative operator(5x)

First, recall the Constant Multiple Rule:

d over dx derivative operator(cf(x)) = cd over dx derivative operator(f(x))

So, for d over dx derivative operator(f(5x)

We have our constant, c of 5 and our function, f(x) is x

So, d over dx derivative operator(5x) = 5d over dx derivative operator(x)

= 5 (1)

= 5


b) Use the Sum Rule to differentiate:

D(2 + 3x2)

First recall the Sum Rule:

D( f(x) + g(x) = D( f(x)) + D (g(x))

For D( 2 + 3x2)

We have a sum of two functions where:

f(x) = 2 and g(x) = 3x2

Note: 2 can still be a function of x even if there are no x terms in it.

So, subbing these values into the Sum Rule, we get:

D( 2 + 3x2) = D( 2 ) + D (3x2)

= 0 + (2)(3x2 - 1)

= 6x


c) Use the Difference Rule to differentiate:

d over dx derivative operator(7x - 2x2)

First, recall the Difference Rule:

d over dx derivative operator(f(x) - g(x)) = d over dx derivative operator(f(x)) - d over dx derivative operator(g(x))

So for:

d over dx derivative operator (7x - 2x2) we have:

f(x) = 7x and g(x) = 2x2 so:

d over dx derivative operator(7x - 2x2) = d over dx derivative operator(7x) - d over dx derivative operator(2x2)

= 7 - (2)2x2-1

= 7 - 4x


d) Use the Product Rule to differentiate:

D (2x sin(x))

First, recall the Product Rule:?

D (f(x) g(x)) = f '(x) g(x) + f(x) g '(x)

For D (2x sin(x)) we have:

f(x) = 2x and g(x) = sin(x)

Instead of trying to calculate f '(x) and g '(x) all at once, we are less likely to make a mistake if we do it individually.

f '(x) = 2 and g '(x) = cos(x)

Now sub these back into the Product Rule.

D (2x sin(x)) = 2 sin(x) + 2x cos(x)

= 2 sin(x) + 2x cos(x)


e) Use the Quotient Rule to differentiate:

d over dx derivative operatoropening bracket
3x2
cos(x)
closing bracket

First, recall the Quotient Rule:

d over dx derivative operatoropening bracket
f (x)
g (x)
closing bracket
  =   
f '(x) g (x) - f (x) g '(x)
(g (x))2
So for d over dx derivative operatoropening bracket
3x2
cos(x)
closing bracket we have:

f (x) = 3x2 and g (x) = cos(x)

Now we work out their derivatives.

f '(x) = (2)3x2-1 = 6x and g '(x) = -sin(x)

Subbing this into the quotient rule:

d over dx derivative operatoropening bracket
3x2
cos (x)
closing bracket
  =   
6x cos (x) - 3x2 (-sin(x))
(cos (x))2
  =   
6x cos (x) + 3x2sin(x)
cos2 (x)
See More Derivative of Complicated Function Examples

Derivative at a Point

Recalling what a derivative is, a derivative is the equation to a line that gives the slope of the tangent at a point along that line. So if we have a function f(x), then the derivative f’(x) is the equation that will tell us what the gradient or slope is at any point along that line represented by x. If we want to find the slope at a point, we must always calculate the derivative first then input the point, x. If we put the point, x, into the equation for the line f(x), we will get a number, then when we differentiate that number, we will get zero!

Derivative at a point - the tangent at that point

first part of solution to finding the derivative of a fuction, 2x + sin(x) and using that to find the slope at point (1, f(1)) and at (0, f(0))
second part of solution to to finding the derivative of a fuction, 2x + sin(x) and using that to find the slope at point (1, f(1)) and at (0, f(0))