Differentiation of Polynomials
Differentiation of polynomials is the process of finding the gradient function, dy/dx, of a curve made up of terms in powers of x, by multiplying each term by its power and then reducing that power by one. It is the core AQA Level 2 Further Maths calculus skill, used to calculate the gradient of a curve at any given point.
Before you start
Make sure you're comfortable with these topics first:
Method
- Write the equation of the curve as a sum of terms in the form a x^n, expanding any brackets first if needed.
- Differentiate each term separately: multiply the coefficient by the power n, then reduce the power by 1, to get the corresponding term of dy/dx.
- Differentiate any constant term (a number with no x) to 0, since it has zero gradient.
- Add or subtract the differentiated terms in the same order as the original expression to give dy/dx.
- To find the gradient at a specific point, substitute the given x-value into dy/dx and simplify.
- Check the final expression uses the same variable and powers are one less than in the original expression.
Worked example
Find dy/dx for the curve y = 2x^4 - 5x^3 + 6x - 9, and hence find the gradient of the curve at the point where x = -1.
- Differentiate each term: 2x^4 becomes 4 x 2x^3 = 8x^3.
- -5x^3 becomes 3 x -5x^2 = -15x^2.
- 6x becomes 6, and the constant -9 becomes 0.
- So dy/dx = 8x^3 - 15x^2 + 6.
- Substitute x = -1: 8(-1)^3 - 15(-1)^2 + 6 = -8 - 15 + 6.
- Final answer: dy/dx = 8x^3 - 15x^2 + 6, and the gradient at x = -1 is -17.
Practice questions
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Q1Differentiate y = x^6 with respect to x.Show answer
Answer: dy/dx = 6x^5
Q2Find dy/dx for y = 5x^3 - 2x^2 + 7.Show answer
Answer: dy/dx = 15x^2 - 4x
Q3Find dy/dx for y = 3x^4 - x^2 + 8x - 1.Show answer
Answer: dy/dx = 12x^3 - 2x + 8
Q4Expand the brackets for y = (x + 3)(x - 5), then find dy/dx.Show answer
Answer: dy/dx = 2x - 2 (from y = x^2 - 2x - 15)
Q5Find the gradient of the curve y = x^3 - 5x^2 + 6 at the point where x = 4.Show answer
Answer: gradient = 8 (dy/dx = 3x^2 - 10x, then substitute x = 4)
Q6Find the value(s) of x for which the gradient of the curve y = x^3 - 12x is equal to 15.Show answer
Answer: x = 3 or x = -3 (solve 3x^2 - 12 = 15)
Exam-style questions
Written in the style of a GCSE Further Maths exam paper, with a full mark scheme.
A curve has equation y = x^3 + 2x^2 - 4x + 1. Find dy/dx, and calculate the gradient of the curve at the point where x = 1.
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A curve has equation y = x^2 - 4x + 1. Find the equation of the tangent to the curve at the point where x = 6. Give your answer in the form y = mx + c.
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A curve has equation y = x^3 + 3x^2 - 24x + 10. Find the coordinates of the stationary points of the curve and use the second derivative to determine the nature of each.
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See real GCSE Further Maths past-paper questions, with official mark schemes →
Free printable worksheet
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