- Difficulty
- Standard
- Suggested time
- 5 min
- Marks
- 3
Implicit differentiation and inverse trigonometric functions practice question 11
The curve below passes through exactly one point with . Find at that point.
At the point you know but not . How will you find , and should numbers go in before or after you differentiate?
First find the point: set , so and , and solve for .
Differentiate implicitly, using the product rule on both terms, then substitute , .
A fraction, or a decimal to two decimal places.
Theory UBC Math 100 §4.5CLP-1 §2.11Key results
Worked solution
Strategy
Locate the point from the equation, differentiate the relation with product and chain rules, and substitute. Every term built from y picks up a factor y′.
Working
e⁰ = 1 and sin 0 = 0, so the point is (1, 0).
Product rule on x eʸ and on 2x² sin y; the chain rule supplies each y′, short for dy/dx.
sin 0 = 0 removes the third term, and cos 0 = 1 keeps the y′ that sin y contributes.
Answer
The answer is negative one third.
Check
Near (1, 0), with eʸ ≈ 1 + y and sin y ≈ y, the curve is close to x + 3y = 1, a line of slope −1/3.
Where this goes wrong
Differentiating sin y as cos y without the factor y′, which gives −3. y depends on x, so every term built from y picks up dy/dx when it is differentiated.
Leans on
The product rule, from Differentiation rules
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