Differentiate, then substitute: To find the gradient of a curve at x = a: differentiate to get f'(x), then substitute x = a to get the number f'(a).
IB-style question — gradient at a point
Find the gradient of y = x³ − 2x at the point where x = 2.
Step by step
- Differentiate.
- Substitute x = 2.
Final answer
The gradient at x = 2 is 10.
Differentiate first, then plug in: Never substitute before differentiating — you must find f'(x) first, then put in the value.
Set f'(x) equal to the gradient and solve: To find where a curve has a particular gradient m, set f'(x) = m and solve for x. There may be more than one answer.
IB-style question — given gradient
The curve y = x² − 4x has gradient 6 at one point. Find the value of x there.
Step by step
- Differentiate and set equal to 6.
- Solve.
Final answer
x = 5.
A quadratic f' can give two answers: If f'(x) is a quadratic, setting it equal to m may give two x-values — report both.
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Gradient = tan(angle with the x-axis): If a tangent makes an angle θ with the (positive) x-axis, its gradient is tan θ. So set f'(x) = tan θ and solve for x.
IB-style question — given the angle
Find the x-value where the tangent to y = x² makes a 45° angle with the horizontal.
Step by step
- Gradient = tan 45° = 1; set f'(x) equal to it.
- Solve.
Final answer
x = 0.5 (where the gradient is 1).
Convert the angle first: Turn the angle into a gradient with tan, then it's the same 'set f'(x) = m' method.