- Difficulty
- Standard
- Suggested time
- 5 min
- Marks
- 3
Differentiation rules practice question 15
Let . For which does exist?
f′ can exist only where f does. Find the domain of f first: the square root and the denominator each rule out some x.
The derivative of is , which is undefined at . Check that point with the definition of the derivative.
Theory UBC Math 100 §4.1CLP-1 §2.6Key results
Worked solution
Strategy
f′ exists where f is defined and its difference quotient has a finite limit. The quotient rule settles the inside of the domain; the edge x = 0 needs the definition.
Working
The square root needs x ≥ 0; the denominator vanishes at x = ±2, and only 2 is in range.
Quotient rule, then multiply top and bottom by 2√x. This is finite for every x > 0 with x ≠ 2.
At x = 0 the difference quotient has no finite limit, so f′(0) does not exist.
Answer
The answer is f prime exists for every x greater than 0 except x = 2.
Check
Near 0, f(x) ≈ −√x/4, so like √x it has a vertical tangent there: the difference quotient at h = 0.0001 is already about −25.
Where this goes wrong
Including x = 0 because f(0) = 0 is defined. Being defined is not enough: the graph has a vertical tangent at 0, where the difference quotient grows without bound.
Leans on
Differentiability, from The definition of the derivative
Try a similar question
Let . Find .Saved in this browser only.