Questions on differentiating powers of x

Question 1

Differentiate y = x⁵ − 3x² + 7.

Accepted Answer

dy/dx = 5x⁴ − 6x. M1 A1. The constant 7 contributes nothing.

Question 2

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

Accepted 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 does not match the form the question asks for.

Question 3

Differentiate y = 5x^4/3 − 8x^1/4.

Accepted Answer

A rational index goes through the power rule exactly as a whole number does: multiply by the index, then reduce it by 1. The first term gives 5 × (4/3)x^1/3 = (20/3)x^1/3, and the second gives −8 × (¼)x^−3/4 = −2x^−3/4. So dy/dx = (20/3)x^1/3 − 2x^−3/4. M1 for reducing both indices by 1, A1 for the two coefficients. Subtracting 1 from ¼ gives −3/4: the index passes through zero and comes out negative, so the second term can also be written −2/x^3/4.

Question 4

The curve C has equation y = 3x³ − 4x + 1. Find dy/dx, and find the gradient of C at the point where x = −2.

Accepted Answer

Differentiating term by term, dy/dx = 9x² − 4. M1 for the power rule, A1 for the derivative. At x = −2 the gradient is 9 × (−2)² − 4 = 36 − 4 = 32. A1. Square the −2 before multiplying by 9, so that the first term is +36 rather than −36 and the gradient comes out positive. The constant 1 raises the whole curve by 1 without tilting it anywhere, so it never appears in dy/dx.

Question 5

Differentiate y = 4x³ − 2/x.

Accepted 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, so the sign of the term changes.

Question 6

Differentiate y = 6√x + 3/x² − 5x.

Accepted 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. Read 3/x² as 3x⁻²: treating the index as positive reverses the sign of the middle term.

Question 7

Differentiate y = x²(x³ − 4).

Accepted 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.

Question 8

Differentiate y = (2x³ − 5x)/x².

Accepted 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.

Question 9

The curve C has equation y = ax³ + bx, where a and b are constants. The gradient of C is 12 at the point where x = 1, and 39 at the point where x = 2. Find the values of a and b.

Accepted Answer

Differentiating, dy/dx = 3ax² + b, with a and b carried through as constants. M1 for the derivative. The two gradients give 3a + b = 12 and 12a + b = 39. M1 for both equations. Subtracting the first from the second removes b: 9a = 27, so a = 3. A1. Then b = 12 − 9 = 3. A1. As a check, dy/dx = 9x² + 3 is 12 at x = 1 and 39 at x = 2. The information given is about the gradient, so both equations come from dy/dx; substituting into the equation of C instead would give the heights of the two points, which the question says nothing about.

Question 10

Differentiate y = (x + 3)(x − 1)/x for x ≠ 0, giving your answer in the form a + b/x², where a and b are integers.

Accepted Answer

The power rule applies to powers of x, so the product and the division both have to go before it can be used. Expanding the numerator, (x + 3)(x − 1) = x² + 2x − 3. M1 for the expansion. Dividing every term by x gives y = x + 2 − 3x⁻¹. M1 for dividing all three terms, A1 for that form. Differentiating termwise, dy/dx = 1 + 3x⁻², that is 1 + 3/x². A1. The constant 2 produced by the division contributes nothing to the derivative. Bringing the −1 down from x⁻¹ against the −3 in front is what turns the last term positive.

Question 11

Given that y = x⁴ − 6/x, find d²y/dx².

Accepted Answer

Rewrite the fraction as a power first: y = x⁴ − 6x⁻¹. M1 for the rewrite. Then dy/dx = 4x³ + 6x⁻², the second term positive because the index −1 comes down against the −6. A1. Differentiating that in turn, d²y/dx² = 12x² − 12x⁻³, that is 12x² − 12/x³. A1. The second derivative is the same rule applied to the first, and the sign of the reciprocal term alternates as the work goes on, from −6x⁻¹ to +6x⁻² to −12x⁻³, because the index brought down is negative at every step.

Question 12

Given that y = (2√x − 3)²/x for x > 0, show that dy/dx = 6x^−3/2 − 9x⁻². Show that this derivative can also be written as 3(2√x − 3)/x², and hence find the exact value of dy/dx when x = 4.

Accepted 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. M1 for expanding the square, M1 for dividing every term by x, A1 for the printed derivative. To gather the two terms, write each over x²: 6x^−3/2 = 6x^1/2/x² and 9x⁻² = 9/x², so dy/dx = (6√x − 9)/x² = 3(2√x − 3)/x². B1 for that single fraction. At x = 4, √x = 2, so dy/dx = 3(4 − 3)/16 = 3/16. M1 for substituting x = 4, A1 for 3/16. The same value comes from the index form, 6 × 4^−3/2 − 9 × 4⁻² = 6/8 − 9/16 = 3/16. 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.

Question 13

A curve has equation y = xⁿ for x > 0, where n is a non-zero rational constant. The gradient of the curve at the point where x = 4 is 8 times the gradient at the point where x = 2. Find the value of n, and verify your answer by evaluating both gradients.

Accepted Answer

Differentiating, dy/dx = nxⁿ⁻¹. M1 for the power rule with a general index. The condition states that n × 4ⁿ⁻¹ = 8 × n × 2ⁿ⁻¹. M1 for that equation. Since n is given as non-zero it can be divided out, leaving 4ⁿ⁻¹ = 8 × 2ⁿ⁻¹. Writing 4 as 2² and 8 as 2³ puts both sides in one base: 2²ⁿ⁻² = 2ⁿ⁺², so 2n − 2 = n + 2. dM1 for reducing to a single base and comparing indices, A1 for n = 4. Verifying, y = x⁴ has gradient 4x³, which is 32 at x = 2 and 256 at x = 4, and 256 = 8 × 32. B1 for the check. The condition n ≠ 0 is doing real work: n = 0 gives the horizontal line y = 1, whose gradient is 0 at both points, and 0 = 8 × 0 is true, so without that restriction the value of n would not be unique.

Question 14

The functions f and g are defined for all real values of x by f(x) = x^4/3 and g(x) = x^2/3. Find f′(x) and g′(x), and state for each the values of x at which the derivative exists. Describe what happens to g′(x) as x approaches 0, and state what that says about the tangent to the graph of g at the origin.

Accepted Answer

The power rule handles a rational index in the usual way, so f′(x) = (4/3)x^1/3 and g′(x) = (2/3)x^−1/3, that is 2/(3x^1/3). M1 for reducing each index by 1 and multiplying by it, A1 for f′(x), A1 for g′(x). The cube root is defined for every real number, so f′(x) exists for all values of x, and f′(0) = 0. In g′(x) the cube root sits in a denominator, so g′(x) exists for x ≠ 0. B1 for both sets. As x approaches 0 the denominator 3x^1/3 approaches 0 and |g′(x)| increases without bound: it is 20/3 at x = 0.001 and 200/3 at x = 0.000001. The graph of g therefore becomes arbitrarily steep near the origin, and its tangent there is vertical. B1 for the unbounded gradient with the vertical tangent. Both functions have a value at x = 0. Being defined at a point and being differentiable there are separate conditions, and only f satisfies the second.

Question 15

The function f is defined by f(x) = (x³ − 8)/(x − 2) for x ≠ 2. Express f(x) as a polynomial, find f′(x), and find the value of f′(3). State, with a reason, whether f′(2) exists.

Accepted Answer

Factorise the numerator as a difference of two cubes: x³ − 8 = (x − 2)(x² + 2x + 4). M1 for the factorisation. Because x ≠ 2 the factor (x − 2) cancels, leaving f(x) = x² + 2x + 4 for x ≠ 2. A1. Differentiating the polynomial, f′(x) = 2x + 2, again for x ≠ 2. M1 for the power rule, A1 for the derivative. At x = 3, f′(3) = 6 + 2 = 8. A1. f′(2) does not exist, because f has no value at x = 2: a derivative is the limit of chord gradients from the point, and no chord can start where the function is undefined. B1 for the reason. The formula 2x + 2 returns 6 at x = 2, which is the gradient the graph would have if the missing point were filled in, and that is a different claim. Cancelling a factor changes the domain, so a cancelled expression carries its restriction with it.