- Difficulty
- Challenging
- Suggested time
- 6 min
- Marks
- 4
Implicit differentiation and inverse trigonometric functions practice question 25
For the circle below, find at a point where , in terms of and .
A second derivative means differentiating once more. When you do, is a constant, or does it still depend on ?
First . Differentiate that quotient with respect to , remembering that depends on : its derivative is , not 0.
Substitute , put everything over , and use .
A formula in x and y. 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 §4.5CLP-1 §2.11Key results
Worked solution
Strategy
Differentiate the first derivative again, treating y as a function of x, then substitute the first derivative and simplify with the equation of the circle.
Working
Implicit differentiation of the circle.
The quotient rule on x/y; the derivative of the y underneath is y′.
Substitute y′ = −x/y, then multiply top and bottom by y.
On the circle, x² + y² = 25.
Negative on the upper half: the top of the circle bends downward.
Answer
The answer is the second derivative equals negative 25 over y cubed.
Check
Near the top, y = √(25 − x²) ≈ 5 − x²/10, so y″(0) = −1/5, and the formula gives −25/5³ = −1/5.
Where this goes wrong
Treating y as a constant when differentiating −x/y, which gives −1/y. y changes with x, so the quotient rule brings in y′, and substituting y′ = −x/y is what produces the x² term.
Leans on
The quotient rule, from Differentiation rules
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