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

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

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

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

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

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.

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.

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)

D

f(x)

g(x)

=

g(x)D( f(x) ) – f(x)D( g(x))

(g(x))^{2}

=

gD(f) – fD(g)

g^{2}

or

=

g(x)f '(x) – f(x)g '(x)

g^{2}

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

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)

Differentiate the following by using the given rule.

a) Use the Constant Multiple Rule to differentiate (5x)

First, recall the Constant Multiple Rule:

(cf(x)) = c(f(x))

So, for (f(5x)

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

So, (5x) = 5(x)

= 5 (1)

= 5

b) Use the Sum Rule to differentiate:

D(2 + 3x^{2})

First recall the Sum Rule:

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

For D( 2 + 3x^{2})

We have a sum of two functions where:

f(x) = 2 and g(x) = 3x^{2}

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 + 3x^{2}) = D( 2 ) + D (3x^{2})

= 0 + (2)(3x^{2 - 1})

= 6x

c) Use the Difference Rule to differentiate:

(7x - 2x^{2})

First, recall the Difference Rule:

(f(x) - g(x)) = (f(x)) - (g(x))

So for:

(7x - 2x^{2}) we have:

f(x) = 7x and g(x) = 2x^{2} so:

(7x - 2x^{2}) = (7x) - (2x^{2})

= 7 - (2)2x^{2-1}

= 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:

3x^{2}

cos(x)

First, recall the Quotient Rule:

f (x)

g (x)

=

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

(g (x))^{2}

So for

3x^{2}

cos(x)

we have:

f (x) = 3x^{2} and g (x) = cos(x)

Now we work out their derivatives.

f '(x) = (2)3x^{2-1} = 6x and g '(x) = -sin(x)

Subbing this into the quotient rule:

3x^{2}

cos (x)

=

6x cos (x) - 3x^{2} (-sin(x))

(cos (x))^{2}

=

6x cos (x) + 3x^{2}sin(x)

cos^{2} (x)

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!