Differentiation rules, question 24: UBC MATH 100 practice

Question 24 of 26Challenging4 marks · 7 minTheory UBC Math 100 §4.1Key results

Prove the power rule $\dfrac{d}{dx}x^{n} = nx^{n-1}$ for every whole number $n \geq 1$ , in two parts. First, show from the definition of the derivative that it holds for $n = 1$ . Then show that if it holds for some whole number $n$ , the product rule makes it hold for $n + 1$ . Explain why the two parts together cover every whole number.

Write your reasoning in full, then compare it against the model solution.

Your answer

Theory UBC Math 100 §4.1Key results

Worked solution

Strategy

Prove the first case from the definition, then show how the product rule passes the rule from one power to the next. A chain of such steps reaches every whole number.

Working

$\frac{d}{dx}x = \lim_{h \to 0} \frac{(x + h) - x}{h} = \lim_{h \to 0} \frac{h}{h} = 1 = 1 \cdot x^{0}$
The case n = 1: the quotient equals 1 for every h ≠ 0, so its limit is 1.
$x^{n+1} = x \cdot x^{n}$
Split one factor of x off the higher power.
$\frac{d}{dx}x^{n+1} = 1 \cdot x^{n} + x \cdot nx^{n-1} = x^{n} + nx^{n} = (n + 1)x^{n}$
The product rule, with (x)′ = 1 and the assumed (xⁿ)′ = nxⁿ⁻¹. The result is the power rule for n + 1.
The first part gives n = 1. The step then turns n = 1 into n = 2, n = 2 into n = 3, and so on without end, so the rule holds for every whole number n ≥ 1.

Answer

\frac{d}{dx}x^{n} = nx^{n-1} \text{ for every whole number } n \geq 1

What earns the marks

For n = 1, writes the difference quotient of x: ((x + h) − x)/h = h/h = 1 for every h ≠ 0, so the limit is 1, which is 1 · x⁰.
For the step, writes xⁿ⁺¹ as the product x · xⁿ and applies the product rule, using the derivative 1 of x and the assumed derivative nxⁿ⁻¹ of xⁿ.
Simplifies 1 · xⁿ + x · nxⁿ⁻¹ to xⁿ + nxⁿ = (n + 1)xⁿ, which is the power rule with n + 1 in place of n.
Explains that the case n = 1 gives n = 2, that gives n = 3, and so on, so every whole number is reached one step at a time.

Check

The step with n = 2 gives 1 · x² + x · 2x = 3x², which matches (x³)′ found from the definition by expanding (x + h)³.

Where this goes wrong

Checking the rule for a few powers, such as x² and x³, and stopping there. Examples cannot cover every n; the step from n to n + 1 is what reaches all of them.