Maths › Differentiation › Differentiating powers of x

Differentiating powers of x

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.

Year 12-13EDEXCEL 9MA0 7.2

Builds on The derivative from first principles and Indices and surds.

IN THIS TOPIC

  • Differentiate xn for any rational n, with sums, differences and constant multiples.
  • Rewrite roots, reciprocals and quotients as powers before differentiating.
  • Expand or simplify products and fractions that hide differentiable powers.

WHAT YOU PROBABLY THINK

To differentiate a product, differentiate each factor.

The power rule

First principles delivered 2x from x2 and 3x2 from x3, and the pattern they suggest holds for every rational power, on the must-learn list as

y = xn ⇒ dydx = nxn−1NOT IN THE BOOKLET — LEARN IT

with two companions that make it an assembly line: a constant multiple rides along, and sums and differences are differentiated term by term. A constant on its own differentiates to zero, having no slope to report.

Short tangent segments drawn on y equals x squared at x equal to minus 1, 0, 1 and 1.5, with slopes minus 2, 0, 2 and 3: the slope at every point is twice the x value, so the gradient is itself the function 2xslope −2slope 0slope 2slope 3the slope readout is itself a function: 2x
FIG. 1The rule, seen: tangent slopes on y = x² at four points read −2, 0, 2 and 3, always twice the x-value. The slope readout is the function 2x.

WORKED EXAMPLE

Three terms, three exponent styles

Differentiate y = 4x3 − 6√x + 2/x2.

Rewrite everything as powers first: y = 4x3 − 6x1/2 + 2x−2.

Apply the rule termwise: dy/dx = 12x2 − 3x−1/2 − 4x−3.

In the original notation, dy/dx = 12x2 − 3/√x − 4/x3.

The rewriting step is where the marks live. Fractional and negative exponents obey the same rule as whole ones, but only once they are visibly exponents.

The curve y equals the square root of x with its tangent at the point 4 comma 2: the tangent's slope is one quarter, exactly what the power rule gives for x to the half at x equal to 4y = √xslope ¼ at (4, 2)fractional powers obey the same rule
FIG. 2The rule on a fractional power: y = √x is x to the ½, so its tangent at (4, 2) has slope ¼, exactly ½ times 4 to the power −½, drawn and computed.

Rewrite, then differentiate

The rule sees only powers of x, so products and quotients must be rewritten before it can act, expanded if a product, split termwise if a fraction. This is also where the opening lie meets its counter example. Differentiating each factor of (2x + 5)(x − 1) would give 2 × 1 = 2, but the true derivative, found by expanding first, is nothing like it.

YOUR TURN

Expand first

Differentiate y = (2x + 5)(x − 1), before opening the working.

Show the working

Expand: y = 2x2 + 3x − 5.

Differentiate termwise: dy/dx = 4x + 3.

At x = 1 this is 7, while the factor-by-factor fake gives 2 everywhere. One numeric spot-check at any point exposes the lie, and a correct rule for products arrives in Year 13.

TRY IT UNSEEN

Split the fraction

Differentiate y = (x2 + 3x − 5)/(4√x).

Show the working

Divide each term by 4x1/2: y = ¼x3/2 + ¾x1/2 − (5/4)x−1/2.

Differentiate termwise: dy/dx = (3/8)x1/2 + (3/8)x−1/2 + (5/8)x−3/2.

Tidied, dy/dx = (3√x)/8 + 3/(8√x) + 5/(8x√x).

No quotient was ever differentiated, only powers. Reshaping the expression until the power rule applies is the whole craft at this stage.

THE EXAM BIT

  • Rewrite before differentiating: every root, reciprocal and quotient becomes x to a power, and the rewriting line is usually a mark of its own.
  • Apply nxn−1 with the sign of n kept: the derivative of x−2 is −2x−3, negative exponent growing more negative.
  • Constants vanish; constant multiples survive unchanged in front.
  • Products must be expanded and fractions split, term by term, before the rule touches them.
  • Present final answers in the notation of the question, converting x−1/2 back to 1/√x where the question used surds.

CHECK YOURSELF

Differentiate y = x4 − 2/√x + 7, and evaluate the gradient at x = 1.

Show a hint

Rewrite 2/√x as a power first; the 7 contributes nothing.

Show the answer

Rewritten, y = x4 − 2x−1/2 + 7, so dy/dx = 4x3 + x−3/2.

That is 4x3 + 1/(x√x) in surd form.

At x = 1 the gradient is 4 + 1 = 5, and the constant 7 never appeared, exactly as it should not.

Multiply down by the exponent, then knock the exponent down by one: nx to the n minus 1.

The rule sees only powers, so rewrite products, roots and fractions until powers are all there is.

CHECK YOUR PROGRESS

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  • Differentiate xn for any rational n, with sums, differences and constant multiples.
  • Rewrite roots, reciprocals and quotients as powers before differentiating.
  • Expand or simplify products and fractions that hide differentiable powers.

No animated video for this topic yet; these notes stand alone.