Mathematics for Management -- Supplementary Electronic Materials

Quiz: Implicit Differentiation

After having worked through the electronic materials (set of notes and video clips), please self-evaluate your learning progress by completing the following quiz. Therefore, please check the correct answer of each of the given questions (single choice) and press the submit button after you completed all 10 questions. (Sometimes the page needs a reload such that the grading script starts to work.)

Your selections will persist so that you can review your answers in case you did not achieve the complete grade. At the bottom of the page you can also find the solution key.

1.The slope of the curve \(y^3 - xy^2 = 4\) at the point where \(y = 2\) is

\( -2 \) \( \frac{1}{4} \) \( -\frac{1}{2} \) \( \frac{1}{2} \)

2. The slope of the curve \(y^2 - xy - 3x = 1\) at the point \((0, -1)\) is

\( -1 \) \( -2 \) \( 1 \) \( 2 \)

3. The equation \(y^2 + 4x^2 = 3xy + x + 9\) defines \(y\) implicitly as a function of \(x\) near the point \((2,1)\). Determine the value of \(y' = \frac{\textrm{d} y}{\textrm{d} x}\) at \(x = 2\) and \(y = 1\).

\( 1 \) \( 2 \) \( 3 \) \( 4 \)

4. The equation of the line tangent to the hyperbola \(x^2 - y^2 = 12\) at the point \((4,2)\) is

\( x - 2y + 6 = 0 \) \( y = 2x \) \( y = 2x - 6 \) \( x + 2y = 6 \)

5. The line tangent to the curve \(y^2 - xy + 9 = 0\) is vertical when

\( y = 0 \) \( y = \pm \sqrt{3} \) \( y = \frac{1}{2} \) \( y = \pm 3 \)

6. If a point moves on the curve \(x^2 + y^2 = 25\), then what value does the second-order derivative \(y'' = \frac{\textrm{d}^2 y}{\textrm{d} x^2}\) at the point \((0,5)\).

\( 0 \) \( \frac{1}{5} \) \( -5 \) \( -\frac{1}{5} \)

7. Use implicit differentiation to determine \(y'\) for the curve \(x^3 - xy + y^3 = 1\).

\( y' = \frac{3x^2}{x - 3y^2} \) \( y' = \frac{3x^2 - 1}{1 - 3y^2} \) \( y' = \frac{y - 3x^2}{3y^2 - x} \) \( y' = \frac{3x^2 + 3y^2 - y}{x} \)

8.Use implicit differentiation to determine \(y'\) for the curve \(x + \cos(x+y) = 0\).

\( y' = \frac{x}{\sin(x+y)} \) \( y' = \csc(x+y) - 1 \) \( y' = \frac{1}{\sqrt{1-x^2}} \) \( y' = \csc(x+y) \)

9. Use implicit differentiation to determine \(y'\) for the curve \(\sin(x) - \cos(y) - 2 = 0\).

\( y' = -\cot(x) \) \( y' = -\frac{\cos(x)}{\sin(y)} \) \( y' = -\csc(y) \cos(x) \) \( y' = \frac{2-\cos(x)}{\sin(y)} \)

10. Use implicit differentiation to determine \(y'\) for the curve \(3x^2 - 2xy + 5y^2 = 1\).

\( y' = \frac{3x + y}{x - 5y} \) \( y' = \frac{y - 3x}{5y - x} \) \( y' = \frac{3x + 4y}{x} \) \( y' = 3x + 5y \)

Your grade is: __ of 10

In case you cannot find the correct answers even after reviewing the electronic materials again, please click on the "+" sign to read more.

Solution: 1d; 2a; 3c; 4b; 5d; 6a; 7c; 8d; 9b; 10b

Here, a, b, c, d indicate the 1st, 2nd, 3rd, and 4th answer choice, respectively, for the numbered questions.