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Each term is differentiated according to the formula:

y = ax^b --> dy/dx = b*ax^(b-1)

Constants differentiate to zero.

Therefore in this...

Remember that the gradient of a curve is expressed as dy/dx. This can be solved by using the chain rule:

dy/dx = dy/dt*dt/dx. The dt in the denominator of the first term, and the numerato...

When solving simultaneous equations you are aiming to use both equations together to find out the value of x and y for values that with ‘fit’ in each equation....

This is an A* grade extension GCSE maths question on rearranging equations.

The question is asking us to find an equation that would represent x. In the question...

Rearranging the terms of the series into the usual "descending order" for polynomials, we get a series expansion of:  

ax^{n-1 +}........ax + a

A basic pro...