# A-level Mathematics/OCR/C1/Differentiation

< A-level Mathematics‎ | OCR‎ | C1

## Motivation

Finding the gradient of a straight line is simple. For a line $y=mx + c$, the gradient is m. But how do you find the gradient of a curve at a particular point? Suppose we want to find the gradient of the tangent line to $y=x^2$ at the point (3,9). Immediately, this question poses a problem. How are we supposed to find the gradient of a line when only one point can be known? We can use a sequence of chords:

x y m
4 16 7
3.75 14.0625 6.75
3.5 12.25 6.5
3.25 10.5625 6.25
3.005 9.030025 6.01

The gradient of the tangent line at (3, 9) is 6. As you can see, this method is rather tedious. Luckily for us we have the derivative, which can accurately and quickly determine any rate of change.

## Derivative

The derivative is the rate of change of a function. It is found using the gradient of a line formula when the distance between the first and second point approaches 0: $\lim_{h \to 0}\frac{f\left(x+h\right)-f\left(x\right)}{h} = f'(x)$. A function is differentiable at the point a only if the limit exists at the point a. The notation for derivative is $\frac{dy}{dx}\ or\ f' \left ( x \right )$.

### Differentiation Rules

1. Derivative of a constant function:

$\frac{d}{dx} \left (c \right) = 0$

2. The Power Rule:

$\frac{d}{dx} \left (x^n \right) = nx^{n - 1}$

3. The Constant Multiple Rule:

If c is a constant and f(x) is a differentiable function: $\frac{d}{dx} c f \left ( x \right ) = c \frac{d}{dx} f \left ( x \right )$

4. The Sum Rule:

$\frac{d}{dx} \begin{bmatrix} f \left ( x \right ) + g \left ( x \right ) \end{bmatrix} = \frac{d}{dx} f \left ( x \right ) + \frac{d}{dx} g \left ( x \right )$

5. The Difference Rule:

$\frac{d}{dx} \begin{bmatrix} f \left ( x \right ) - g \left ( x \right ) \end{bmatrix} = \frac{d}{dx} f \left ( x \right ) - \frac{d}{dx} g \left ( x \right )$

The derivative is used to find the gradient of a line at the point a. Then the point a is substituted into the derivative. The resultant will give us the gradient at the point a. The first derivative of a distance function is velocity and the second derivative is acceleration.

The distance that a car has travelled is given by the function $s = 1.25t^2 -5$t. Where s is the distance travelled in metres and t is the time elapsed. What is the velocity of the car when 12 seconds have elapsed?

1) Using the power, the constant multiple rule and the difference derivative rules we get v = 2.5t - 5:

2) Now we substitute the time into the function: v = 2.5x12 - 5, we get that v = 25 metres/second.

So after 12 seconds the car will be travelling at 25 metres per second.

### Tangent Lines

Since we can find the gradient of a function at any point we can also find the equation of the tangent line to the function at any point:

What is the equation of the tangent line for the curve $y = x^2 + 3x -16$ at the point x = -4?

First we need to find the derivative of the equation.

$\frac{dy}{dx} = 2x + 3$

Then we find the gradient at the point x = -4.

$2\left ( -4 \right) + 3 = -5$

Using the original function we find the corresponding y value to the x value at the point of tangency.

$y = -4^2 + 3 \left (-4 \right) -16 = -12$

Then we use the point-gradient equation to find the equation.

$y + 12 = -5 \left ( x + 4 \right )$ so the equation of the tangent line is y = -5x - 32....

### Normal Lines

A Normal line is the line perpendicular to the Tangent Line. Therefore when we determine the equation of the normal line the procedure is very similar to when we determining the equation of the tangent line. Example:

What is the equation of the normal line for the curve $y = x^2 + 3x -16$ at the point when x=-4?

First we need to find the derivative of the equation.

$\frac{dy}{dx} = 2x + 3$

Then we need to solve for the given x value.

$2\left ( -4 \right) + 3 = -5$

Then we use the perpendicular line equation to find the gradient of the normal line.

$m_1 \times m_2=-1$ so $-5 \times m_2=-1$ so the gradient is $m_2=\frac{-1}{-5}$ which is $\frac{1}{5}$.

Then we find the y value corresponding to x value

$y = -4^2 + 3 \left (-4 \right) -16 = -12$

Then we use the point-gradient equation to find the equation.

$y + 12 = .2 \left ( x + 4 \right )$ so the equation of the normal line is y = .2x - 11.2

## Higher Derivatives

By C2, you will be expected to have a brief understanding of the applications of the second (and third, in some cases) derivative of a function. The second derivative is exactly what it sounds like, it is the derivative of a derivative. The third derivative is the derivative of the second derivative, and so on.The notation for the second derivative is, assuming we are talking in terms of y with respect to x: $\frac{d^2y}{dx^2}$ And subsequent derivatives simply have larger numbers in their indices.

## Application of Derivatives to Graphs

Derivatives also help us to graph functions, by locating minimum, maximum, and inflection points. They can also determine the interval on which a function is convex or concave.

Convex - The graph is below the tangent lines. Example $-x^2$, think of a frown.

Concave - The graph is above the tangent lines. Example $x^2$, think of a smile.

Inflection Point - The point when a function changes from convex to concave or vice versa. Example $x^3$ at x = 0.

Maximum Point - The highest point of a function on a convex interval.

Minimum Point - The lowest point of a function on a concave interval.

Stationary Point - A point where f'(c) = 0

### Rules of Stationary Points

• If $f' \left ( c \right ) = 0$ and $f'' \left ( c \right ) <0$, then c is a local maximum point of f(x). The graph of f(x) will be convex on the interval.
• If $f' \left ( c \right ) = 0$ and $f'' \left ( c \right ) >0$, then c is a local minimum point of f(x). The graph of f(x) will be concave on the interval.
• If $f' \left ( c \right ) = 0$ and $f'' \left ( c \right ) = 0$ and $f''' \left ( c \right ) \ne 0$, then c is a local inflection point of f(x).

### Locating And Evaluating Stationary Points

The Function

The original function is used to determine when the function crosses the x and y axis's.

The First Derivative Test

The first derivative is used to find all the minimum, maximum or inflection points. At these points f'(x) = 0. Also if we make a sign chart we can see at which intervals the function is increasing and decreasing, but the second derivative is a better test of this.

The Second Derivative Test

The second derivative is used to find the points when a function is concave or when it is convex at these points f''(x) = 0. Then you need to make a sign chart. The interval(s) that the sign of the second derivative is positive the function is concave and if the sign is negative the function is convex. When f'(a) is zero and on that interval f''(x) is negative then the point a is a maximum point, before this point the function will be increasing afterwards the function will be decreasing. When f'(a) is zero and on that interval f''(x) is positive then the point a is a minimum point, before this the function will be decreasing afterwards the function will be increasing.

The Third Derivative Test

The third derivative test determine if a point is an inflection point. This usually is not necessary If f''(c) = 0 and $f'''(c) \ne 0$ the f(c) is an inflection point.

#### An Example

Sketch the graph of $f(x) = x^3 -4x^2$

1) The function will tell us that:

$f(x) = x^2(x - 4)$
$0 = x^2(x - 4)$ x = 0 and x=4 are the x-intercepts.
$y = 0^2(0 - 4)$ y = 0 is the y-intercept.

2) Then we use the first derivative test.

$f'(x) = 3x^2 - 8x$
$0 = x(3x - 8)$ x = 0 and x=2.66 are the minimum or maximum or inflection points.

3) Then we use the second derivative test.

f''(x) = 6x - 8
0 = 6x -8 so x = $\frac{4}{3}$.
We now make a number line __f'''(0) = -8______ 1.333____f'''(2) = 4.
Before 1.333 the function will be convex after 1.333 the function will be concave.

4) Then we use the third derivative test.

f'''(x) = 6
$f'''(\frac{4}{3}) = 6$ The point is an inflection point.

5) Now we gather all the information into a table. Also we need to find the corresponding y values to all the x values that we obtained from steps 1 to 3 using f(x).

Interval or Point f(x) behaviour
x < 0 Increasing
(0,0) Maximum Point, x-intercept, and y-intercept
0 < x < 1.33 Decreasing
(1.33,-4.74) Inflection Point
1.33 < x < 2.66 Decreasing
(2.66,-9.48) Minimum Point
2.66 < x < $\infty$ Increasing
(4,0) x-intercept

6)Now we can draw our graph:

This is part of the C1 (Core Mathematics 1) module of the A-level Mathematics text.