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Differentiation with Graphs

GCSELevel 8-9Cambridge iGCSEEdexcel iGCSE

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} = 2

So, the gradient, or rate of change, of this line is 2.


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

y = 2x + 1\\

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

Level 8-9GCSECambridge iGCSEEdexcel iGCSE

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).


\dfrac{\text{d}y}{\text{d}x} = 2x

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

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

Therefore, 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

Level 8-9GCSECambridge iGCSEEdexcel iGCSE

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).

\dfrac{\text{d}y}{\text{d}x} = 4x


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

4x = 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)

Level 8-9GCSECambridge iGCSEEdexcel iGCSE

Example 1: Gradients of Lines

Find the gradients of the 3 lines on this graph:

[3 marks]

y = -2x + 10\\ \dfrac{\text{d}y}{\text{d}x} = -2


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


y = x + 2\\

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

Level 8-9GCSECambridge iGCSEEdexcel iGCSE

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:

y=3x^2+12x+1\\ \dfrac{\text{d}y}{\text{d}x}=6x+12


At the stationary point, the gradient will be 0.

\dfrac{\text{d}y}{\text{d}x}=6x+12=0\\ 6x=-12\\ x=-2


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.

Level 8-9GCSECambridge iGCSEEdexcel iGCSE

Differentiation with Graphs Example Questions



The gradient of the line is 4.




Stationary point occurs when the gradient is 0:

-2x+4=0\\ 2x=4\\ x=2\\

Sub x into the curve equation:


y =10


This is a maximum point.


\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 =1

Stationary 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)

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