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Differentiation (International GCSE)
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Differentiation finds the gradient of a curve at any point. It is part of the International GCSE Higher tier.
The rule
If y = axn, then dydx = naxn−1
Multiply by the power, then reduce the power by 1.
y = 4x3 → dydx = 12x2
The derivative of a number on its own is 0.
Multiply by the power, then reduce the power by 1.
y = 4x3 → dydx = 12x2
The derivative of a number on its own is 0.
Gradient at a point
Substitute the x-value into dydx.
y = x2 + 3x: dydx = 2x + 3. At x = 2, gradient = 7.
y = x2 + 3x: dydx = 2x + 3. At x = 2, gradient = 7.
Turning points
At a turning point the gradient is 0. Solve dydx = 0.
y = x2 − 6x + 1: 2x − 6 = 0, so x = 3 and y = −8.
In kinematics, velocity is dsdt and acceleration is dvdt.
y = x2 − 6x + 1: 2x − 6 = 0, so x = 3 and y = −8.
In kinematics, velocity is dsdt and acceleration is dvdt.
Find the gradient of y = 2x3 − 5x at x = 1.
- dydx = 6x2 − 5
- At x = 1: 6 − 5 = 1
Answer: 1
Multiply by the power and reduce the power by 1; constants disappear. Substitute to get a gradient. Set dydx = 0 to find turning points.
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