- Difficulty
- Standard
- Suggested time
- 5 min
- Marks
- 3
The definition of the derivative practice question 20
Let . Write the difference quotient as a single fraction in and , with the factor cancelled.
Here h is not 0, and neither x nor x + h is −3.
This asks for the quotient itself, so stop before any limit. How do you turn a difference of two fractions into one fraction?
. Subtract over the common denominator .
Expand the numerator . Everything cancels except a multiple of .
A formula in x and h, simplified so that it is defined when h = 0. 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
Substitute x + h into f, subtract f(x) over a common denominator, and expand the numerator. It reduces to a multiple of h, which cancels with the h underneath.
Working
Put both fractions over the common denominator.
Expand both products; only 3h survives.
Divide by h and cancel it.
Letting h → 0 now would give the derivative; the question stops one step earlier.
Answer
The answer is 3 over the product of x + 3 and x + h + 3.
Check
At x = 0 and h = 1 the formula gives 3/(3·4) = 1/4, which is f(1) − f(0) = 1/4 − 0 directly.
Where this goes wrong
Letting h go to 0 and giving f′(x) = 3/(x + 3)². The difference quotient is the expression before the limit, so the h in x + h + 3 stays.
Leans on
Combining fractions over a common denominator, from Functions, graphs and inverses
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