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Practice Set A — Differentiation: Composite, Implicit, and Inverse Functions

20 questions • 30 minutes • auto-graded with full solutions
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Auto-graded practice from the Differentiation: Composite, Implicit, and Inverse Functions question bank.
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Multiple Choice
Question 1
If 2x + 3y = 6, what is dy/dx?
Solution: Differentiate implicitly: 2 + 3(dy/dx) = 0, so dy/dx = -2/3 (the slope of the line).
Question 2
If g is the inverse of f, which formula gives g'(a)?
Solution: The inverse-function derivative rule states (f^(-1))'(a) = 1 / f'(f^(-1)(a)), i.e., g'(a) = 1 / f'(g(a)).
Question 3
If f(x) = 2x + 3, what is the derivative of its inverse function?
Solution: The inverse is f^(-1)(x) = (x - 3)/2, whose derivative is 1/2. (Equivalently, 1/f'(x) = 1/2.)
Question 4
What is d/dx [ arcsin(x) ]?
Solution: The derivative of arcsin(x) is 1 / sqrt(1 - x^2).
Question 5
What is d/dx [ arctan(x) ]?
Solution: The derivative of arctan(x) is 1 / (1 + x^2).
Question 6
What is d/dx [ arccos(x) ]?
Solution: The derivative of arccos(x) is -1 / sqrt(1 - x^2).
Question 7
If f(x) = x^4, what is f''(x)?
Solution: f'(x) = 4x^3, and f''(x) = 12x^2.
Question 8
If f(x) = sin(x), what is f''(x)?
Solution: f'(x) = cos(x), and f''(x) = -sin(x).
Question 9
If f(x) = x^3 - 2x, what is f''(x)?
Solution: f'(x) = 3x^2 - 2, and f''(x) = 6x.
Question 10
What is d/dx [ 5 arctan(x) ]?
Solution: The constant multiple carries through: 5 * 1/(1 + x^2) = 5/(1 + x^2).
Question 11
For the circle x^2 + y^2 = 25, what is the slope of the tangent line at the point (3, 4)?
Solution: dy/dx = -x/y. At (3, 4), the slope is -3/4.
Question 12
If sin(y) = x, what is dy/dx?
Solution: Differentiate implicitly: cos(y)(dy/dx) = 1, so dy/dx = 1/cos(y).
Question 13
If e^y = x (with x > 0), what is dy/dx?
Solution: Differentiate implicitly: e^y(dy/dx) = 1, so dy/dx = 1/e^y. Since e^y = x, dy/dx = 1/x.
Question 14
If x^2 - y^2 = 1, what is dy/dx (assuming y not 0)?
Solution: Differentiate implicitly: 2x - 2y(dy/dx) = 0. Solving gives dy/dx = x/y.
Question 15
On the circle x^2 + y^2 = 16, at which points is the tangent line horizontal?
Solution: dy/dx = -x/y = 0 requires x = 0. This occurs at (0, 4) and (0, -4), the top and bottom of the circle.
Question 16
If x = cos(y), what is dy/dx?
Solution: Differentiate implicitly: 1 = -sin(y)(dy/dx), so dy/dx = -1/sin(y).
Question 17
For y^2 = x^3 (a semicubical parabola), what is dy/dx at the point (1, 1)?
Solution: Differentiate: 2y(dy/dx) = 3x^2, so dy/dx = 3x^2/(2y). At (1, 1): 3(1)/(2(1)) = 3/2.
Question 18
The equation of the tangent line to x^2 + y^2 = 25 at (3, 4) is:
Solution: The slope at (3, 4) is dy/dx = -x/y = -3/4. Using point-slope form: y - 4 = -3/4 (x - 3).
Question 19
If f(2) = 5 and f'(2) = 3, and g is the inverse of f, what is g'(5)?
Solution: g'(5) = 1 / f'(g(5)) = 1 / f'(2) = 1/3, since g(5) = 2.
Question 20
Let f(x) = x^3. If g is the inverse of f, what is g'(8)?
Solution: g(8) = 2 since 2^3 = 8. f'(x) = 3x^2, so f'(2) = 12. Then g'(8) = 1/f'(2) = 1/12.