Gradient at a point by differentiation revision notes

Edexcel International GCSE Maths (4MA1), chapter 28: Calculus, specification point 3.4c. Free lesson.

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Differentiation gives a rule for the gradient at any x-value. Use it to find a gradient at a point, or set it equal to a given gradient to find the points where that steepness occurs.

Find the gradient at a given x-value

The derivative, written as dy/dx, is a rule for the gradient that still contains x. Differentiate the curve first, then substitute the x-value. This works because the derivative gives the gradient at any point on the curve. For y = x³ − 4x² + 5, the derivative is dy/dx = 3x² − 8x. The constant +5 disappears. To find the gradient at x = 3, substitute 3 into the derivative, using brackets: 3(3)² − 8(3) = 27 − 24 = 3.

Why you must differentiate first

Substituting into the original equation first turns the curve equation into a single number. A number has derivative 0, so this would not give the gradient of the original curve. Keep x in the derivative until you are ready to evaluate it. The correct order is differentiate first, then substitute the x-value. Brackets around the value you substitute help prevent sign errors, especially with negative values.

Find points with a given gradient

When a question gives you a gradient, set the derivative equal to that number. Setting it equal to 0 only finds places where the curve is flat, such as stationary or turning points. For y = x³ + 3x² − 24x, the derivative is 3x² + 6x − 24. To find where the gradient is 21, solve 3x² + 6x − 24 = 21. This gives 3x² + 6x − 45 = 0, then 3(x + 5)(x − 3) = 0. The x-values are −5 and 3. Substitute each into the original equation to find the matching y-values.

Give coordinates, not just x-values

The x-values show where the curve has the required gradient, but the question asks for points. Use the original equation to find the y-coordinate for each x-value. For y = x³ + 3x² − 24x, x = −5 gives y = 70, and x = 3 gives y = −18. The coordinates are (−5, 70) and (3, −18). Both points count.

Check the method on a short example

For y = x² − 5x, differentiate to get 2x − 5. At x = 4, the gradient is 2(4) − 5 = 3. A rate of change is another way a question may describe the gradient.