- Difficulty
- Standard
- Suggested time
- 5 min
- Marks
- 3
The definition of the derivative practice question 18
Use the definition of the derivative to find for the function below.
Rules such as the power rule earn no marks on this question.
The numerator of the difference quotient is a difference of two fractions. What turns it into something with a visible factor of ?
Combine over the common denominator .
The numerator becomes . Expand the square in full, factor out , and cancel it before letting .
A formula in x. Use ^ for powers and brackets for every input: sin(2x), e^(x^2), log(x+1). Give exact values: 1/6, not 0.167.
Theory UBC Math 100 §3.3CLP-1 §2.2Key results
Worked solution
Strategy
Set up the difference quotient, combine the two fractions so the factor h in the numerator becomes visible, cancel it, and only then let h approach 0.
Working
Put the two fractions over the common denominator x²(x + h)².
Expand the square in full; the x² terms cancel and h factors out.
Cancel h. Nothing is left that blows up as h → 0.
Let h approach 0, then simplify.
Answer
The answer is f prime of x equals negative 2 over x cubed.
Check
At x = 1 the formula gives −2, and f(1.01) = 1/1.0201 ≈ 0.9803, a drop of about 0.0197 over a step of 0.01.
Where this goes wrong
Expanding (x + h)² as x² + h², which loses the 2xh term and makes the answer collapse to 0. Write the square out in full: x² + 2xh + h².
Leans on
Combining fractions over a common denominator, from Functions, graphs and inverses
Try a similar question
Use the definition of the derivative to find for the function below.Saved in this browser only.