The definition of the derivative is nice to prove why certain derivatives are the way they are, but as you have experienced they can be tedious at times. Therefore, let me state the short cut differentiation rules that we can use from now on.

RULE 1:

EXAMPLE 1:

Find the derivative of f (x) = 6.

SOLUTION: f ' (x) = 0

RULE 2:

EXAMPLE 2:

Find the derivative of f (x) = x ⁴.

SOLUTION: f ' (x) = 4x ^{4 - 1} = 4x ³

EXAMPLE 3:

Find the derivative of f (x) = x.

SOLUTION: f (x) = x = x ¹

f ' (x) = 1x ^{1 - 1} = 1x ⁰ = 1

EXAMPLE 4:

Find the derivative of f (x) = x ³.

SOLUTION: f ' (x) = 3x ^{3 - 1} = 3x ²

RULE 3:

If u is a differentiable function of x, and c is a constant, then

What this is saying is that we can pull the constant out in front of the function, and then take the derivative of the variable.

EXAMPLE 5:

Find the derivative of f (x) = 6x ⁴.

SOLUTION: f ' (x) = 6(4x ^{4 - 1}) = 24x ³

EXAMPLE 6:

Find the derivative of f (x) = 7x ⁹.

SOLUTION: f ' (x) = 7(9x ^{9 - 1}) = 63x ⁸

RULE 4:

If u and v are differentiable functions of x, then their sum/difference u ± v is differentiable at every point where u and v are both differentiable. As such points,

EXAMPLE 7:

Find the derivative of f (x) = 3x - 4.

SOLUTION: f ' (x) = 3(1x ^{1 - 1}) - 0 = 3x ⁰ = 3

EXAMPLE 8:

Find the derivative of f (x) = x ³ - 4x ² + 3x + 2.

SOLUTION: f ' (x) = 3x ^{3 - 1} - 4(2x ^{2 - 1}) + 3(1x ^{1 - 1}) + 0 = 3x ² - 8x + 3

EXAMPLE 9:

Find the derivative of f (x) = 6x ⁴ - 3x ³ + 6x ² - 2x + 4.

SOLUTION:

f ' (x) = 6(4x ^{4 - 1}) - 3(3x ^{3 - 1}) + 6(2x ^{2 - 1}) - 2(1x ^{1 - 1}) + 0 = 24x ³ - 9x ² + 12x - 2

As we have just seen, the derivative of a sum is the sum of the derivatives. Unfortunately, the derivative of a product is not the product of the derivatives.

RULE 5:

If u and v are differentiable at x, then so is their product uv, and

EXAMPLE 10:

Find the derivative of f (x) = (2x + 3)(3x - 4).

SOLUTION:

EXAMPLE 11:

Find the derivative of f (x) = (6x ² + 3x - 1)(x ³ + 3x ² - 3x + 1).

SOLUTION:

RULE 6:

If u and v are differentiable at x, and v (x) ¹ 0, then the quotient u/v is differentiable at x, and

EXAMPLE 12:

SOLUTION:

EXAMPLE 13:

SOLUTION:

EXAMPLE 14:

SOLUTION: We will have to use both the quotient rule and product rule to find this derivative.

RULE 7:

In fact, if n is a rational number, then the power rule will still apply.

EXAMPLE 15:

SOLUTION: First of all, I am going to rewrite the expression using negative exponents.

Now I will find the derivative.

EXAMPLE 16:

Find the derivative of y = (x + x ^-1)(x - x ^-1 + 1).

EXAMPLE 17:

Find the derivative of y = 2(x ^{- 1/ 2} + x ^{1/ 2}).

So far we have been finding the first derivative of a function, but we can find other order derivatives using the same rules. Let us discuss the notation for these higher order derivatives.

FIRST ORDER DERIVATIVE

SECOND ORDER DERIVATIVE

THIRD ORDER DERIVATIVE

Nth ORDER DERIVATIVE

EXAMPLE 18:

Find the first and second derivative for the following function.

SOLUTION: I will first simplify this function first before I find the derivatives.

EXAMPLE 19:

Find the first and second derivatives of the following function.

SOLUTION: I will first simplify this function before I find the first and second derivatives.

EXAMPLE 20:

Find the derivatives of all orders for the following function.

EXAMPLE 21:

(A) Find the equations for the horizontal tangents to the curve y = x ³ - 3x - 2. Also find the equations for the lines that are perpendicular o these tangents at the points of tangency. (B) What is the smallest slope on the curve? At what point on the curve does the curve have this slope.

SOLUTION: (A) The points where the curve has horizontal tangents, the slope is zero. Furthermore, the first derivative gives us the slope of the tangent line at any point on the curve. So find the first derivative and then set it equal to zero and solve for x.

The two points on the curve that have horizontal tangents are (-1, 0) and (1, -4).

These tangent lines have slope of 0.

So the equation of the tangent line through the point (-1, 0) is y = 0.

The equation of the tangent line through the point (1, -4) is y = -4.

The perpendicular line to a horizontal tangent line is a vertical line.

The perpendicular line to y = 0 is x = -1.

The perpendicular line to y = -4 is x = 1.

(B) What is the smallest slope on the curve? First of all, y' = 3x ² - 3 gives the slope of the tangent line at any point on the curve. So, to find the smallest slope on the curve, I will find the second derivative. Then determine where it is equal to zero.

y'' = 3(2x ^{2 - 1}) - 0 = 6x = 0 ® x = 0

When x = 0, y = -2, and the point (0, -2) is the point where the slope is the smallest.

Let us look at the graphs of the original function and the first two derivatives to see if we can find a relationship between them.

Do you see a connection between the function and the first derivative, and the first derivative and the second derivative?

The first derivative gives us the minimum and maximum points of the original function. The second derivative gives us the minimum and maximum points of the first derivative.

I just wanted to touch on this topic, but we will discuss it in depth later on in this course.