Find and repair a rule error
Testing a claimed derivative
A student claims that the derivative of \(f(x) = x\,e^x\) is \(e^x\).
The orange tangent shows the true slope of the curve. In the lower graph, the chalk curve gives the true slope at every x, and the orange bar marks its value at the point. The red curve is the slope that the claim predicts.
Go to \(x = 1\). The tangent has slope \(2e\), about 5.44, but the red curve is at \(e\), about 2.72, so one mismatch shows that the claim is wrong.
Go to \(x = -1\), where the curve has its lowest point. The tangent is flat there, but the claim predicts \(e^{-1}\), about 0.37.
A derivative must match the slope everywhere
A claimed derivative predicts the slope of the curve at every x. In the lower graph, the red curve is the prediction and the chalk curve is the true slope.
A correct derivative matches the slope at every x, so one mismatch is enough to show that a claim is wrong.
Drag the point along the curve. The red and chalk curves meet at one x although the claim is wrong, so one match does not prove a claim.
Notation
Name the last operation, the one you would do last to compute \(f(x)\). For \(x\,e^x\), it is a product, with this rule.
\[(uv)' = u'v + uv'\]Compare the claim with the rule piece by piece: the number of terms, the inside slopes, and the signs.
A product has two terms. A composition multiplies by the slope of the inside. The quotient numerator is \(u'v - uv'\), in that order.
Write the repaired derivative. Then compare it with the slope of the tangent at one \(x\).
\[\cy{f'(x)} \overset{?}{=} \text{slope of the tangent}\]Test at an \(x\) where no term of the rule is 0. A wrong or missing piece that is 0 there does not show in the test.
When it fails
At \(x = 0\), the claim \(e^x\) and the true slope are both 1. The missing term \(x\,e^x\) is 0 there, so the test cannot see it.
One match does not prove a claim, so confirm a repair at a second \(x\).
Example 1: a sign error
A student writes that the derivative of \(f(x) = \dfrac{x}{x+2}\) is \(-\dfrac{2}{(x+2)^2}\). Find the error and repair it.
Test the claim first. For \(x > -2\) the curve rises, so the true slope is positive. The claim is negative, so it is wrong.
The last operation is a quotient, with top \(u = x\) and bottom \(v = x + 2\). The numerator is \(u'v - uv'\).
\[f'(x) = \dfrac{1\cdot(x+2) - x\cdot 1}{(x+2)^2}\]The student wrote the terms in the other order, which reverses the sign. The repaired derivative is positive.
\[f'(x) = \dfrac{2}{(x+2)^2}\]At \(x = 0\), the repaired slope is \(\tfrac12\), the slope of the orange tangent. The claim gave \(-\tfrac12\).
\[f'(0) = \dfrac{2}{4} = \dfrac12\]
Example 2: a missing inside slope
A student writes that the derivative of \(f(x) = \ln(1 + x^2)\) is \(\dfrac{1}{1+x^2}\). Find the error and repair it.
Test the claim at \(x = 1\). The orange tangent has slope 1, but the claim gives \(\tfrac12\), so the claim is wrong.
The last operation is a logarithm of \(1 + x^2\), so the chain rule applies. The rule divides the slope of the inside by the inside, and the claim leaves out the inside slope \(2x\).
\[(\ln u)' = \dfrac{u'}{u}\]A tempting wrong move: Repair the claim by doubling it.
\[\dfrac{2}{1+x^2}\]At \(x = 1\) this gives 1, so it passes the test. At \(x = 0\) it gives 2, but the curve has its lowest point there and the tangent is flat.
Use the inside slope \(2x\), not a guessed factor. This gives 0 at \(x = 0\), where the tangent is flat, and 1 at \(x = 1\).
\[f'(x) = \dfrac{2x}{1 + x^2}\]
Practice
A student writes that the derivative of \(h(x) = x\ln x\), for \(x > 0\), is \(\ln x + \dfrac1x\). Find the error and repair it.
Do not test at x = 1, because the claim and the true slope agree there. Test at x = 2 and read the orange tangent.
Solution
The last operation is a product with \(u = x\) and \(v = \ln x\). The rule is \(u'v + uv'\).
\[h'(x) = 1\cdot\ln x + x\cdot\dfrac1x\]The student wrote \(\tfrac1x\) for the second term, but the second term is \(x\cdot\tfrac1x = 1\).
\[h'(x) = \ln x + 1\]As a check at \(x = 2\), the repair gives \(\ln 2 + 1 \approx 1.693\), the slope of the orange tangent. The claim gives \(\ln 2 + \tfrac12 \approx 1.193\).
Questions
The figure shows \(f(x) = x\,e^x\) and the claim that \(f'(x) = e^x\). The red curve is the claimed slope.
At \(x = -1\) the tangent is flat and the red curve is at about 0.37. What does this show?
- The claim is wrong, because the true slope is 0 and the claim is not.
- The claim is correct, because \(e^x\) is positive everywhere.
- The claim is correct, because 0.37 is close to 0.
Answer
The claim is wrong, because the true slope is 0 and the claim is not.
At \(x = 0\) the orange slope and the red curve are both at 1. What does this show?
- Nothing yet. The missing term \(x\,e^x\) is 0 at \(x = 0\), so the test must use another \(x\).
- The claim is correct, because the two curves meet.
- The claim is wrong, because the curves should differ at every \(x\).
Answer
Nothing yet. The missing term \(x\,e^x\) is 0 at \(x = 0\), so the test must use another \(x\).
A student writes the derivative of \(y = \dfrac{3x}{x^2+4}\) as \(\dfrac{3(x^2+4) - 3x\cdot 2x}{x^2+4}\). Which expression is the repaired derivative?
- \(\dfrac{12 - 3x^2}{(x^2+4)^2}\)
- \(\dfrac{12 - 3x^2}{x^2+4}\)
- \(\dfrac{3x^2 - 12}{(x^2+4)^2}\)
- \(\dfrac{12 + 9x^2}{(x^2+4)^2}\)
Answer
\(\dfrac{12 - 3x^2}{(x^2+4)^2}\)
Summary
You test a claimed derivative against the slope of the curve, then repair it with the rule for the last operation.
This works because a correct derivative gives the slope of the tangent at every \(x\), so one mismatch shows an error.
Use it on your own derivatives, at an \(x\) where no term of the rule is 0.
This is one of four skills in topic 3.5 of Which & Why: AP Calculus.