A derivative gives the gradient of the curve at a point, so you can test it at one value of x. Work out what your formula says at that x, then estimate the gradient from two nearby points on the original function. If the two numbers agree closely, the derivative is very likely correct.
This lesson checks the work from powers, products, the chain rule and ln and ex forms. It uses a calculator, so it is a revision habit rather than a method for non-calculator questions.
How do you check a derivative at a point?
- Choose a convenient x, such as 1 or 2, where the arithmetic is simple.
- Evaluate your derivative at that x.
- Choose a small step, h = 0.01 is enough.
- Estimate the gradient: (f(x + h) − f(x − h)) / (2h).
- Compare. They should match to about three significant figures.
Using x − h and x + h, rather than x and x + h, makes the estimate more accurate for the same effort.
Worked example
A student differentiates y = x(2x − 1)³ and gets dy/dx = (2x − 1)²(8x − 1). Check the answer at x = 1.
Step 1, formula value: at x = 1, (2 − 1)² × (8 − 1) = 1 × 7 = 7.
Step 2, function values with h = 0.01:
- f(1.01) = 1.01 × (1.02)³ = 1.01 × 1.061208 = 1.07182008.
- f(0.99) = 0.99 × (0.98)³ = 0.99 × 0.941192 = 0.93178008.
Step 3, estimate: (1.07182008 − 0.93178008) / 0.02 = 0.14004 / 0.02 = 7.002.
The formula gives 7 and the estimate gives 7.002. They agree, so the derivative passes.
The mistake to watch for
Now suppose a different student forgot the inner derivative and wrote dy/dx = (2x − 1)³ + 3x(2x − 1)².
Mistaken value at x = 1: 1 + 3 = 4
Estimate from the function: 7.002
The gap of about 3 is far larger than rounding error, so the check fails. It also points to where to look: the second term should contain an extra factor of 2 from the inner derivative, giving 6x(2x − 1)², which at x = 1 is 6. Then 1 + 6 = 7.
Do not trust a check that agrees only at x = 0 or x = 1 when your formula contains an x term. At those values a factor of x can hide a mistake, so also test at x = 2.
Check yourself
1. A student says that the derivative of y = √x is 1/(2√x). Check at x = 4 using h = 0.01, given √4.01 ≈ 2.00250 and √3.99 ≈ 1.99750.
Show answer
Formula: 1/(2 × 2) = 0.25.
Estimate: (2.00250 − 1.99750) / 0.02 = 0.00500 / 0.02 = 0.25. They agree, so the derivative passes.
2. Check that the derivative of y = x³ at x = 2 is 12, using h = 0.1.
Show answer
f(2.1) = 9.261 and f(1.9) = 6.859.
Estimate: (9.261 − 6.859) / 0.2 = 2.402 / 0.2 = 12.01. This agrees with 3 × 2² = 12. ✓
3. A student claims that the derivative of y = ln(x² + 4) is 1/(x² + 4). At x = 2 with h = 0.01, ln(8.0401) − ln(7.9601) ≈ 0.0100. Does the claim pass?
Show answer
Formula value: 1/8 = 0.125.
Estimate: 0.0100 / 0.02 = 0.5.
It fails. The correct derivative is 2x/(x² + 4), which at x = 2 is 4/8 = 0.5, matching the estimate.
Where this leads next
Work through the differentiation techniques practice set and use this check on a few answers. The calculus shape and rate explorer shows the same idea visually, with a tangent that you can compare against the curve. When the derivative is trusted, you can move on to stationary points and tangents and normals.
Some students check every answer and run out of time, while others never check. A teacher in online one-to-one Additional Mathematics tuition can help you decide which answers deserve a check. The non-calculator working trainer covers the exact-arithmetic side.