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Differentiating powers of x questions

First principles proves the pattern; this lesson industrialises it. One rule differentiates every power of x, whole, negative or fractional, and with sums and constant multiples it handles any polynomial in sight, provided the expression is first rewritten into powers the rule can see.

7 original questions · 20 marks · the differentiating powers of x notes · Differentiation

Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.

• 2 marks · warm-up

  Differentiate y = x⁵ − 3x² + 7.

  Worked answer

  dy/dx = 5x⁴ − 6x. M1 A1. The constant 7 contributes nothing. A flat line has gradient zero, wherever it sits.

• 2 marks · warm-up

  Differentiate y = √x, giving your answer without fractional indices.

  Worked answer

  Rewrite as x^1/2. The power rule then gives (1/2)x^−1/2, which is 1/(2√x). M1 for the rewrite and the power rule, A1 for 1/(2√x). The rewrite comes first, because the rule cannot see a root sign. Leaving the answer as ½x^−1/2 ignores the instruction and drops the final accuracy mark.

• 3 marks

  Differentiate y = 4x³ − 2/x.

  Worked answer

  Rewrite the fraction as −2x⁻¹. Then dy/dx = 12x² + 2x⁻² = 12x² + 2/x². M1 for rewriting as a power, A1 for 12x², A1 for the second term. Bringing down the −1 multiplies the −2 to give +2, and that positive sign is what the question exists to test. An answer ending in −2/x² loses the accuracy mark.

• 3 marks

  Differentiate y = 6√x + 3/x² − 5x.

  Worked answer

  Rewrite as 6x^1/2 + 3x⁻² − 5x, then apply the rule term by term. That gives 3x^−1/2 − 6x⁻³ − 5, or 3/√x − 6/x³ − 5 in the original notation. M1 for rewriting all three terms as powers, A1 A1 for the derivative. Every term has to become a power of x before the rule touches it. The commonest error is reading 3/x² as if the index were positive, which flips the sign of the middle term.

• 2 marks

  Differentiate y = x²(x³ − 4).

  Worked answer

  Expand first, so y = x⁵ − 4x² and dy/dx = 5x⁴ − 8x. M1 for expanding, A1 for the derivative. The power rule handles sums, not products. Differentiating each bracket and multiplying gives 2x × 3x² = 6x³, which is wrong.

• 2 marks

  Differentiate y = (2x³ − 5x)/x².

  Worked answer

  Divide through first. y = 2x − 5/x = 2x − 5x⁻¹, so dy/dx = 2 + 5x⁻² = 2 + 5/x². M1 for dividing through, A1 for the derivative. A quotient with a single-term denominator splits by division, and no quotient rule is needed here.

• 6 marks · stretch

  Given that y = (2√x − 3)²/x for x > 0, show that dy/dx = 6x^−3/2 − 9x⁻². Hence find the exact value of x at which the curve has a stationary point.

  Worked answer

  Expand the square before dividing. (2√x − 3)² = 4x − 12√x + 9, so dividing every term by x gives y = 4 − 12x^−1/2 + 9x⁻¹. Differentiating gives dy/dx = 6x^−3/2 − 9x⁻², as required. Setting that equal to zero, 6x^−3/2 = 9x⁻². Multiply both sides by x² to get 6x^1/2 = 9, so √x = 3/2 and x = 9/4. M1 for expanding the square, M1 for dividing every term by x, A1 for the printed derivative, M1 for setting the derivative to zero, dM1 for clearing the negative indices, A1 for x = 9/4. Three separate traps sit in this one question. Squaring the bracket as 4x + 9 loses the middle term, dividing only the first term by x wrecks the indices, and −1/2 − 1 has to come out as −3/2 rather than −1/2.

The same practice on paper: the printable workbook for this topic, questions and a worked answer book.

Practise differentiating powers of x one question at a time

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