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Differentiating powers of \(x\), including negative and fractional powers, with sums, differences and constant multiples. This appears in almost every calculus question, so it must be automatic.
Key facts
\(x^n \to nx^{n-1}\)
Notes
Powers of \(x\)
\(\dfrac{\mathrm{d}}{\mathrm{d}x}(x^n) = nx^{n-1}\) for any rational \(n\). A constant differentiates to 0.
Rewrite roots and fractions as powers first: \(\sqrt{x} = x^{\frac{1}{2}}\),\(\dfrac{1}{x^3} = x^{-3}\), \(\dfrac{3}{2\sqrt{x}} = \dfrac{3}{2}x^{-\frac{1}{2}}\)
Expand brackets, and split fractions with a single term on the bottom: \(\dfrac{x^3 - 4}{x^2} = x - 4x^{-2}\)
Differentiate sums, differences and constant multiples term by term. For example, \(\dfrac{\mathrm{d}}{\mathrm{d}x}(kx) = k\) and \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{1}{x}\right) = -\dfrac{1}{x^2}\)
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own.
Differentiate expressions with powers of \(x\)
3 to 4 marksGrade D
Rewrite every term as a multiple of a power of \(x\)
Multiply by the power, then reduce the power by 1.
Tidy up; only write negative or fractional powers back as roots or fractions if asked.
Example. Given that \(y = 4x^3 - 6\sqrt{x} + \dfrac{5}{x^2}\),\(x > 0\), find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
M1 for \(x^{n} \to x^{n - 1}\) for at least one term
A1 for two terms correct (may be unsimplified)
A1 for \(\frac{\mathrm{d}y}{\mathrm{d}x} = -15x^{4} - 2 - 2x^{-\frac{3}{2}}\), fully correct and simplified
Worked solution: In index form, \(y = -3x^{5} - 2x + 4x^{-\frac{1}{2}}\). \(\frac{\mathrm{d}y}{\mathrm{d}x} = -15x^{4} - 2 - 2x^{-\frac{3}{2}} = -15x^{4} - 2 - \frac{2}{x\sqrt{x}}\)
Question 2Medium6 marks
The curve \(C\) has equation \(y = \frac{(x^{2} - 1)^{2}}{x^{3}}\), \(x \gt 0\).
(a) Find \(\frac{\mathrm{d}y}{\mathrm{d}x}\), giving your answer in the form \(A + Bx^{-2} + Cx^{-4}\), where \(A\), \(B\) and \(C\) are constants.[4]
(b) Hence find the exact value of the gradient of the curve \(y = \frac{(x^{2} - 1)^{2}}{x^{3}}\) at the point where \(x = 2\).[2]
Show the answer and mark scheme
(a)Answer: \(\frac{\mathrm{d}y}{\mathrm{d}x} = 1 + 2x^{-2} - 3x^{-4}\) \((A = 1, \ B = 2, \ C = -3)\)
M1 for writing \(y\) as a sum of powers of \(x\), e.g. by expanding the numerator, \(x^{4} - 2x^{2} + 1\), and dividing each term by \(x^{3}\)
A1 for \(y = x - 2x^{-1} + x^{-3}\)
M1 for \(x^{n} \to x^{n - 1}\) for at least one of their terms
A1 for \(\frac{\mathrm{d}y}{\mathrm{d}x} = 1 + 2x^{-2} - 3x^{-4}\), i.e. \(A = 1, \ B = 2, \ C = -3\)
Worked solution: By expanding the numerator, \(x^{4} - 2x^{2} + 1\), and dividing each term by \(x^{3}\): \(y = x - 2x^{-1} + x^{-3}\). So \(\frac{\mathrm{d}y}{\mathrm{d}x} = 1 + 2x^{-2} - 3x^{-4}\), giving \(A = 1, \ B = 2, \ C = -3\).
(b)Answer: \(\frac{21}{16}\)
M1 for substituting \(x = 2\) into their \(\frac{\mathrm{d}y}{\mathrm{d}x}\)
A1 for \(\frac{21}{16}\) (cao)
Worked solution: At \(x = 2\): \(\frac{\mathrm{d}y}{\mathrm{d}x} = 1 + 2(2)^{-2} - 3(2)^{-4} = \frac{21}{16}\).