Differentiation rules practice question 7
A student claims that for all differentiable functions and . She tries and , finds that both sides are , and concludes that the claim is true. Which response is correct?
The claim is about every pair of differentiable functions. What would it take to prove that, and what would it take to show it false?
One pair where the two sides differ, even at a single , is enough to sink the claim. Try the simplest functions that are not constant.
Theory UBC Math 100 §4.1Key results
Worked solution
Strategy
A claim about all functions is disproved by a single counterexample: one pair f, g and one x where the two sides differ. Agreement on one pair proves nothing, because the claim is about every pair.
Working
The left side of the claim, worked out directly.
The right side of the claim for the same pair.
The two sides differ, so the claim is false. The product rule gives the true derivative: (fg)′ = f′g + fg′ = x + x = 2x.
With constants both sides are 0, so her example agrees with the claim. That shows only that the claim survives one test; a rule about all functions needs a proof, and one failure is enough to refute it.
Answer
The answer is The claim is false: for f(x) = g(x) = x the derivative of the product is 2x, while the product of the derivatives is 1. One agreeing example does not prove a rule..
Check
The product rule gives (x · x)′ = 1 · x + x · 1 = 2x, the same as differentiating x² directly, and 2x is not the constant 1.
Where this goes wrong
Taking one agreeing example as proof. A rule about every pair of functions can be refuted by one counterexample but never proved by examples, however many agree.
The idea behind it
The differentiation rules lesson, a figure to move, in a few steps
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