- Difficulty
- Standard
- Suggested time
- 5 min
- Marks
- 3
Implicit differentiation and inverse trigonometric functions practice question 14
Find the -intercept of the line tangent to the curve at the point where .
To find an intercept of the tangent line you first need the line: a point and a slope at . Which rule does the slope of need?
Product rule: the derivative of is . Evaluate it, and the curve, at .
Write the tangent line as , set , and solve for .
An exact expression, or a decimal to three decimal places.
Theory UBC Math 100 §4.7CLP-1 §2.12Key results
Worked solution
Strategy
Find the point of tangency and the slope there (product rule, with the derivative of arctan), write the tangent line in point-slope form, and set y = 0.
Working
arctan 1 = π/4 because tan(π/4) = 1.
Product rule: both x and arctan x depend on x.
Point-slope form at the point of tangency.
Set y = 0 and solve for x.
Answer
The answer is 2 over (pi plus 2), about 0.389.
Check
The tangent line has height −½ at x = 0 and π/4 at x = 1, so it crosses the x-axis between 0 and 1, as 0.389 does.
Where this goes wrong
Differentiating x arctan x as x/(1 + x²) or as arctan x alone. Both factors depend on x, so the product rule gives two terms and the slope at 1 is π/4 + 1/2.
Leans on
Equation of a tangent line, from The definition of the derivative
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