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AQA Core 1 Differentiation and Integration Revision Notes
Differentiation:

Rates of Change
· The gradient of a curve is defined as the gradient of the tangent
Gradient is denoted as dy/dx if y is given as a function of x
Gradient is denoted by f ’(x) if the function is given as f(x)
· The process of finding dy/dx or f’(x) is known as DIFFERENTIATING
Derivatives:
If f(x) = x^n then f '(x) = nx^n-1 (^ mean to the power of)

If f(x) = a (where a is any number) then f '(x) = 0
Example:

y = x3 + 4x2 – 3x + 6

then dy/dx = 3x2 + 8x – 3

Using Differentiation

 If the value of dy/dx is positive at x = a, then y is increasing at x = a
 If the value of dy/dx is negative at x = a, then y is decreasing at x = a
 Points where dy/dx= 0 are called stationary points
To find the coordinates of a stationary point solve dy/dx= 0, then substitute
into y=f(x) to find y values
...

However always make sure
you write + C in your
answer

The area under a graph can be found using integration
The area under the graph of y=f(x) between x=a and x=b is found by
evaluating the definite integral
e
...
find the area under the curve of y=4x-x^3 between the lines x=0 and x=2

Remember: an area below the x axis has a NEGATIVE VALUE



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