# Differentiation with Graphs

## Differentiation with Graphs Revision

**Differentiation with Graphs**

**Differentiation **allows us to find the variable rate of change, so when looking at a graph, differentiation gives us the gradient at any given point.

Make sure you are happy with the following topic before continuing:

**Gradients of Lines**

The following graph shows the line y = 2x + 1

We know to find the **gradient** we can find 2 coordinates on the line and calculate \dfrac{\text{change in} y}{\text{change in} x}:

Using (0,1) and (3, 7):

\dfrac{7-1}{3-0} = 2So, the **gradient**, or rate of change, of this line is 2.

We could also calculate this by** differentiating** the equation of the line:

\dfrac{\text{d}y}{\text{d}x}=1\times 2x^{1-1} + 0 = 2

**Gradients of Curves**

We can also use **differentiation** to find the **gradients** of curves at given points.

The following graph shows the curve y = x^2 + 2

Curves have many **gradients**, depending on which point you are looking at.

For example:

Find the **gradient** of the curve y = x^2 + 2 at the point (2,6).

**Differentiate**:

Sub in the value of x from (2,6):

\dfrac{\text{d}y}{\text{d}x} = 2(2) = 4Therefore, the **gradient** at the point (2,6) is 4.

If we were asked the same question but with the point (4, 18), we would substitute x=4 into the derivative:

\dfrac{\text{d}y}{\text{d}x} = 2(4) = 8

**Stationary Points**

A stationary point on a curve occurs when the **gradient** is 0.

The curve below has 2 stationary points, one that is a **maximum** point, and one that is a **minimum** point.

The **gradient** at both the maximum and minimum point is 0.

**Finding the Coordinates of Stationary points:**

For example, if a curve had the equation y = 2x^2 + 4, we can use **differentiation** to find the coordinates of the stationary point(s).

At a stationary point, we know the **gradient**, 4x is 0:

We solve this equation:

x = 0

Therefore, the stationary point occurs when x=0. To find the full coordinate, we can sub this x coordinate back into the curve’s equation:

y = 2(0^2) + 4 = 4

Coordinate: (0,4)

**Example 1: Gradients of Lines**

Find the **gradients** of the 3 lines on this graph:

**[3 marks]**

y = 3x - 3\\ \dfrac{\text{d}y}{\text{d}x} = 3

y = x + 2\\

\dfrac{\text{d}y}{\text{d}x} = 1

**Example 2: Stationary Points**

Find the coordinates of the stationary point of the following curve, y = 3x^2+12x+1, deciding whether this is a minimum or maximum point.

**[4 marks]**

We can **differentiate** to find the **gradient**:

At the stationary point, the **gradient** will be 0.

Sub this into the equation to find the y coordinate:

y=3(-2)^2+12(-2)+1\\ y=-11

Coordinate: (-2,11)

We can see from the graph that this is a minimum point, as the curve decreases to this point, and then increases after.

## Differentiation with Graphs Example Questions

**Question 1: **Find the gradient of the line y=4x+3

**[2 marks]**

Differentiate:

\dfrac{\text{d}y}{\text{d}x}=4The gradient of the line is 4.

**Question 2: **The graph below shows the curve y=-x^2+4x+6.

Find the coordinate of the stationary point, stating whether this is a minimum or maximum point.

**[3 marks]**

Differentiate:

\dfrac{\text{d}y}{\text{d}x}=-2x+4

Stationary point occurs when the gradient is 0:

-2x+4=0\\ 2x=4\\ x=2\\Sub x into the curve equation:

y=-(2)^2+4(2)+6

y =10

(2,10)

This is a maximum point.

**Question 3: **Find the coordinates of the 2 stationary points of the curve y=\dfrac{1}{3}x^3+2x^2+3x+1

**[5 marks]**

Differentiate:

\dfrac{\text{d}y}{\text{d}x} = x^2+4x+3

Gradient will be 0 at the stationary points, so:

x^2+4x+3=0\\ (x+3)(x+1)=0\\x=-3 and x=-1

When x=-3,

y=\dfrac{1}{3}(-3)^3+2(-3)^2+3(-3)+1 =1Stationary point: (-3, 1)

When x=-1,

y=\dfrac{1}{3}(-1)^3+2(-1)^2+3(-1)+1 =-\dfrac{1}{3}Stationary point: \left(-1, -\dfrac{1}{3}\right)