- Difficulty
- Standard
- Suggested time
- 4 min
- Marks
- 3
Implicit differentiation and inverse trigonometric functions practice question 19
Let for . Find .
This is arcsin of , a composition. What is the outer derivative, and what is the square of the inside?
Chain rule with : .
Here . Now find , the derivative of , and put the pieces together.
A formula in x. 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 §4.7CLP-1 §2.12Key results
Worked solution
Strategy
The derivative of arcsin at the inside √x, times the derivative of √x. Since (√x)² = x, the root in the arcsin derivative becomes √(1 − x).
Working
Outer derivative 1/√(1 − u²) at u = √x, times the inner derivative.
(√x)² = x for x > 0, and the derivative of √x is 1/(2√x).
The same thing under one root, since both factors are positive on 0 < x < 1.
Answer
The answer is f prime of x equals 1 over (2 root x times root (1 minus x)).
Check
Because arcsin √x + arcsin √(1 − x) = π/2, f′(x) must equal f′(1 − x), and 1/(2√x √(1 − x)) is unchanged when x becomes 1 − x.
Where this goes wrong
Stopping at 1/√(1 − x), which leaves out the derivative of the inside. The inside is √x, so the chain rule multiplies by 1/(2√x).
Leans on
The chain rule, from The chain rule and logarithms
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