The process of finding the gradient or slope of a function is the differentiation. We used to find the gradient of a straight line, just by dividing the change in 'y' by change in 'x', in a certain range of values. However, when it comes to a curve, it is not easy to find the gradient or slope as the very thing we want to measure, keeps changing from point to point.


we have to draw tangents at all those points and then find the gradients individually. The following animation illustrates just that.

If we stick to this method we will have to draw hundreds, if not thousands, of tangents to find the gradient at various points of the curve; enough work to put off someone doing maths for decades!

Good news is that there is a method that comes to our rescue. It is differentiation. It provides us with a method that finds the rate of change of graph; the value so obtained is called differential coefficient (dy/dx) - what we call gradient.

If y = xn then dy/dx = nxn-1

That means, if a curve is in the form of y = xn , its gradient at any point is given by nxn-1 . All we have to do is, to substitute 'x' values for the dy/dx and to get the gradient at points.


y = x2 - 2x

So, dy/dx = 2x -2

Now we can find the gradient at any point on the curve by just putting the value of 'x' in it. The animation explains this:


dy/dx =2x -2
dy/dx at x = 0 => dy/dx = 2x0 -2 = -2
dy/dx at x = 1 => dy/dx = 2x1-2 = 0
dy/dx at x = 2 => dy/dx = 2x2 -2 = 2

Just look at the gradients of the tangents; the values from differentiation are compatible with them.


Differentiate the following:

  • y = (x + 3)2
    y = x2 + 6x + 9
    dy/dx = 2x + 6
  • y = x3 - 2x
    dy/dx = 3x2 - 2
  • y = x1/2 + 4x
    dy/dx = 1/2x-1/2 + 4


Sketch the curve y = 4x - x2 and find the equations of tangents at x = 0 and x = 4. The two tangents meet up at A. Find the coordinates of A as well.


dy/dx = 4 - 2x
dy/dx at x =0 => dy/dx = 4
For the tangent,
x = 0; y = 0; m = dy/dx = 4
y = mx + c
0 = 4x0 + c => c = 0
So, the equation of the tangent is y = 4x.
dy/dx at x = 4 => dy/dx = -4
For the tangent,
x = 4; y = 0; m = dy/dx = -4
y = mx + c
0 = 4x-4 + c => c = 16
So, the equation of the tangent is y = -4x + 16.
If two equations cross at A, 4x = -4x + 16
8x = 16
x = 2
y = 2 x 4 = 8
So, the coordinates of A are, (2,8).

Equations of tangents to a curve

tangents to a curve

Differentiation is a very important tool in practical applications. The following example illustrates it.


A farmer wants to make an enclosure for his sheep using a fencing material of length 60m with one side being a wall. He wants to make its shape rectangular so that he can keep as many sheep as possible. How should he choose the dimensions for the enclosure.
Let x be the width. Then the length is (60 - 2x) - only three sides to cover, as fourth is the wall.
If the area is A,
A = x(60 - 2x)
A = 60x - 2x2
Now, let's sketch a graph of A against x.


When the area is maximum, the graph has a peak; differentiation helps us to find the peak - the gradient of the peak is 0
dA/dx = 60 -4x
dA/dx =0 => 60 - 4x = 0
4x = 60
x = 15

So, he should choose width as 15m and length as (60 - 2x15) = 30m, to maximize the area - a brilliant use of differentiation!


Four squares from the corners of a square plate are removed so that it can be turned into a open cubical box. Find the length of a square to be removed in terms of the length of the main square so that the volume of the box is a maximum.


Let the length of the plate be l and that of a small square be x.
If the volume is V,
V = x(l - 2x)2
= x(l2 - 4lx + 4x2)
= l2x - 4lx2 + 4x3
dV/dx = l2 - 8lx + 12x2
At the peak of the graph of V against x, dV/dx = 0
12x2 - 8lx + l2 = 0
x = 8l ± √ 64l2 - 48l2 / 24
x = 8l ± √ 16l2 / 24
x = 8l ± 4l / 24
x = l/2 or l/6
Since l/2 is not possible, x = l/6

Therefore, to maximize the volume, a square of side l/6 should be cut off from each corner of the plate.

Practice is the key to mastering maths; please visit this page, for more worksheets.

Please work out the following questions to complement what you have just learnt.

    1) Sketch y = x2 - x - 6 and find the coordinates of the point with zero gradient. A tangent is drawn for the curve at x = 3. Find its equation.
    2) The product of two numbers is 64. Find the numbers so that their sum is a minimum.
    3) Sketch y = x2 - 3x and find the equation of the tangent at x = 2. Hence find the equation of the perpendicular line drawn at the same point to the curve.
    4) A square sheet is turned into a cubical open box by removing four squares from its vertices. Find the minimum length of a square, so that the volume of the box is maximum.
    5) Find a relationship between the height and diameter of a cylinder when its surface area is maximum.




Recommended Reading


Maths is challenging; so is finding the right book. K A Stroud, in this book, cleverly managed to make all the major topics crystal clear with plenty of examples; popularity of the book speak for itself - 7th edition in print.

Recommended - GCSE & iGCSE


This is the best book available for the new GCSE(9-1) specification and iGCSE: there are plenty of worked examples; a really good collection of problems for practising; every single topic is adequately covered; the topics are organized in a logical order.

Recommended for A Level


This is the best book that can be recommended for the new A Level - Edexcel board: it covers every single topic in detail;lots of worked examples; ample problems for practising; beautifully and clearly presented.