How to use Differentiation to find the gradient at a point

Before reading this post, I would strongly recommend looking at my other post of how to differentiate if you do not know how to do so already.

Assuming you know how to differentiate equations; lets begin.


Here is an example of a common question that involves dy/dx:



First of all breaking down the question it is basically asking you to differentiate the equation, then substitute the value of x to find the gradient at point P.

This is simply done and using simple algebra and substitution we can derive that the gradient has a value of 11.




Very Easy...

We could also be asked this question:


The steps to approaching this would be to:

1) Use the gradient we have already found to find the gradient of the normal.

2) Use substitution to find the value of y

3) Use the new values to find the value of c and complete the equation.



The image below shows using -1/m to find the perpendicular gradient.

Then we use the original formulae of the curve to find the value of y, as shown here -------->










 We then substitute the values of x and y into our new equation in the form of:

y=mx + c

We then find that c = 15.48.

We now have the equation as shown :)

How to Differentiate Equations

Differentiation may seem very confusing at first but it is actually very simple and easy as long as you follow these short instructions.

Differentiation is used to find the gradient in most cases of a graph at a specific point. This is done by finding the dy/dx and the substituting into the new equation an x value to find the gradient at that point.

Here are the short tips to make differentiation easy:


1) Make the equation as simple as you can make it. 


This is done by making the coefficients of x nice numbers and the same with the powers of x. This requires indices rules. For example:




As you can see when given an equation you want to make it as simple as possible. This will make things A LOT easier.

Here is an example of an equation of a line that becomes much more simplified:






2) Multiply the coefficient of x by the power (coefficient means the number before the x!)

This is quite a straightforward step:

In this step you do not want to change the value of  anything except the coefficient of x.

3) Minus 1 from the power!

This can be done as shown:






And that's it! This is an example of an actual question with all the steps:

1)  Original equation



2)  Simplifying values of x


3)  Multiplying the power by the coefficient

Note - anything to the power of 0 = 1

4)  Minus 1 from the powers of x

5)  Easy peasy :)

Resolving Forces - FAST

Here is an object of 2kg lying on a slop raised from the horizontal of an angle of 'a'.
From this we know that there is a force of 2g downwards as a result of (M)x(G).
We also know that the reaction force will be perpendicular to the slope as shown.
Therefore this triangle can be drawn with the bottom line parallel to that of the slope itself.

 We can now look at this triangle in more detail. The angle joining the vertical force of gravity and the reaction force will be the same as the angle of which the slope is raised from the horizontal.

From this we can draw the triangle to the right to show the different forces acting upon the object on the slope.

 As we know the values of the downward force (gravity) and the angle a; we can derive the values of the force parallel to the slope and the value of the reaction force.

The reaction force is worked out using SOHCAHTOA.
Cos = Adjacent/Hypotenuse.
Cos a = Hypotenuse / 2g
Hypotenuse = 2g Cos a

The force parallel to the slope is worked out the same way:
Sin = Opposite / Hypotenuse
Sin a = Opposite / 2g
Opposite = 2g Sin a

The trick to resolving forces is to learn that:

• The force next to the angle will always use 'Cos'.
• The force that is not next to the angle will use 'Sin'.

Therefore we can assume the general formulae:

• 'mg cos a'

for the reaction force of an object.

• 'mg sin a'

for the force parallel to the slope.

Square Any 2 Digit Number In Your Head

Maths Trick - Squaring any 2 digit number - In your head!

An easy and fast maths trick to quickly calculate any number squared in your head!!


1. If your number is between 1 and 9 then you can just learn the answer from memory.

2. Work out what multiple of 10 is closest to your number.
For example 20 is the closest multiple of ten to 22.

3. Calculate the difference between the multiple of ten and your number.

4. Add the difference to your original number.

5. Multiply the new number by the multiple of ten.
For example 20 x 24 = 480.

6. Add the difference squared to this number (2 squared = 4).
480 + 4 = 484.


Example 1.

Square 42.

1. 40 is the closest multiple of ten.
2. The difference is 2.
3. 2 add 42 = 44.
4. 44 x 40 = 1760.
5. The difference squared is 4.
6. 4 add 1760 = 1764.


Example 2.

Square 67.

1. 70 is the closest multiple of ten.
2. The difference is 3.
3. 67 minus 3 = 64.
4. 70 x 64 = 4480.
5. The difference squared is 9.
6. 9 add 4480 = 4489.