- Difficulty
- Standard
- Suggested time
- 4 min
- Marks
- 3
Implicit differentiation and inverse trigonometric functions practice question 9
The point lies on the curve below. Find at that point.
Solving this curve for would be messy, and it has a cross term . How can you find the slope by differentiating the equation as it stands?
Differentiate both sides with respect to , treating as a function of . Every that gets differentiated produces a factor of .
The middle term is a product, so it needs the product rule: .
A decimal or a fraction.
Theory UBC Math 100 §4.5CLP-1 §2.11Key results
Worked solution
Strategy
Differentiate the equation term by term with respect to x, remembering that y depends on x. Then gather the dy/dx terms on one side and solve, substituting the point only at the end.
Working
Differentiate both sides. The right-hand side is a constant, so it becomes 0.
The product rule handles xy; the chain rule turns y² into 2y·dy/dx.
Collect every dy/dx term on one side and factor it out.
Divide. This is the slope at any point on the curve.
Substitute both coordinates. Implicit derivatives normally need x and y.
Answer
The answer is −4/5, or −0.8.
Check
At x = 1.01 the curve’s y is about 1.9920, a drop of about 0.008 from 2 over a step of 0.01, a slope of −0.8.
Where this goes wrong
Differentiating xy as just y, or as just x·dy/dx. It is a product of two functions of x, so both terms appear.
Leans on
The product rule, from Differentiation rules
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