- Difficulty
- Standard
- Suggested time
- 5 min
- Marks
- 3
The definition of the derivative practice question 11
Let . Use the definition of the derivative to find , and then find an equation of the tangent line to at .
Rules such as the chain rule earn no marks for the slope.
There is a square root in the numerator of the difference quotient. Which algebraic move clears a root so that can cancel?
The difference quotient at 2 is . Multiply the top and the bottom by the conjugate .
The tangent line passes through with slope : .
An equation of a line, y = mx + b or y - y0 = m(x - x0). 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
A square root in the difference quotient calls for the conjugate, which turns the numerator into a multiple of h. With the slope found, the point (2, 2) and point-slope form give the line.
Working
Set up the definition at a = 2.
Conjugate, difference of squares, then cancel h.
Let h → 0.
Point-slope form through (2, 2), then solved for y.
Answer
The answer is y equals negative three quarters x plus seven halves.
Check
The line passes through (2, 2) on the curve, and at x = 2.04 it gives 1.97, against f(2.04) = √3.88 ≈ 1.9698.
Where this goes wrong
Writing the difference of squares as h instead of −3h. The numerator is (4 − 3h) − 4 = −3h; dropping the −3 gives slope 1/4 and a line that rises where the curve falls.
Leans on
Rationalising with a conjugate, from Limits
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