BC Calculus Assignment 18: Differentiation Mechanics

For problems 1 - 22, find \(\frac{dy}{dx}\)
  1. \( y = (4x + 1)^2(1 - x)^3 \)
    1. \( (4x + 1)^2(1 - x)^2(5 - 20x) \)
    2. \( (4x + 1)(1 - x)^2(4x + 11) \)
    3. \( 5(4x + 1)(1 - x)^2(1 - 4x) \)
    4. \( (4x + 1)(1 - x)^2(11 - 20x) \)
    5. \( -24(4x + 1)(1 - x)^2 \)
    2. Solution C
  2. \( y = \frac{2-x}{3x + 1} \)
    1. \( -\frac{7}{(3x + 1)^2} \)
    2. \( \frac{6x - 5}{(3x + 1)^2} \)
    3. \( -\frac{9}{(3x + 1)^2} \)
    4. \( \frac{7}{(3x + 1)^2} \)
    5. \( \frac{7 - 6x}{(3x + 1)^2} \)
    3. Solution A
  3. \( y = \sqrt{3 - 2x} \)
    1. \( \frac{1}{2\sqrt{3 - 2x}} \)
    2. \( -\frac{1}{\sqrt{3 - 2x}} \)
    3. \( \frac{(3 - 2x)^{\frac{3}{2}}}{3} \)
    4. \( -\frac{1}{3 - 2x} \)
    5. \( \frac{2}{3} \left( 3 - 2x \right) ^{\frac{3}{2}} \)
    4. Solution B
  4. \( y=\frac{2}{(5x + 1)^3} \)
    1. \( -\frac{30}{(5x + 1)^2} \)
    2. \( -30(5x + 1)^{-4} \)
    3. \( -\frac{6}{(5x + 1)^4} \)
    4. \( -\frac{10}{3}(5x + 1)^{-\frac{4}{3}} \)
    5. \( \frac{30}{(5x + 1)^4} \)
    5. Solution B
  5. \( y= 3x^{\frac{2}{3}} - 4x^{\frac{1}{2}} - 2 \)
    1. \( 2x^{\frac{1}{3}} - 2x^{-\frac{1}{2}} \)
    2. \(3x^{-\frac{1}{3}} - 2x^{-\frac{1}{2}} \)
    3. \( \frac{9}{5}x^{\frac{5}{3}} - 8x^{\frac{3}{2}} \)
    4. \( \frac{2}{x^{\frac{1}{3}}} - \frac{2}{x^{\frac{1}{2}}} - 2 \)
    5. \(2x^{-\frac{1}{3}} - 2x^{-\frac{1}{2}} \)
    6. Solution E
  6. \( y = 2\sqrt{x} - \frac{1}{2\sqrt{x}} \)
    1. \( x + \frac{1}{x\sqrt{x}} \)
    2. \( x^{-\frac{1}{2}} + x^{-\frac{3}{2}} \)
    3. \( \frac{4x - 1}{4x\sqrt{x}} \)
    4. \( \frac{1}{\sqrt{x}} + \frac{1}{4x\sqrt{x}} \)
    5. \( \frac{4}{\sqrt{x}} + \frac{1}{x\sqrt{x}} \)
    7. Solution D
  7. \( y = \sqrt{x^2 + 2x - 1} \)
    1. \( \frac{x + 1}{y} \)
    2. \( 4y(x + 1) \)
    3. \( \frac{1}{2\sqrt{x^2 + 2x - 1}} \)
    4. \( -\frac{x + 1}{ \left( x^2 + 2x - 1 \right) ^{\frac{3}{2}}} \)
    5. None of these
    8. Solution A
  8. \( y=\frac{x}{\sqrt{1 - x^2}} \)
    1. \( \frac{1 - 2x^2}{ \left( 1 - x^2 \right) ^{\frac{3}{2}}} \)
    2. \( \frac{1}{ 1 - x^2} \)
    3. \( \frac{1}{ \sqrt{1 - x^2}} \)
    4. \( \frac{1 - 2x^2}{ \left( 1 - x^2 \right) ^{\frac{1}{2}}} \)
    5. None of these
    9. Solution E
  9. \( \cos {\left( x^2\right) } \)
    1. \( 2x \sin {\left( x^2\right)} \)
    2. \( -\sin {\left( x^2\right)} \)
    3. \( -2\sin{x} \cos{x}\)
    4. \( -2x \sin {\left( x^2\right)} \)
    5. \(\sin{(2x)} \)
    10. Solution D
  10. \( y = \sin^2{(3x)} + \cos^2{(3x)} \)
    1. \( -6sin{(6x)} \)
    2. \( 0 \)
    3. \( 12\sin{(3x)}\cos{(3x)} \)
    4. \( 6(\sin{(3x)}\cos{(3x)}) \)
    5. \( 1 \)
    11. Solution B. This is the function \(y = 1\), whose derivative is everywhere 0.
  11. \( y = \cos^2{x} \)
    1. \( -\sin^2{x} \)
    2. \( 2\sin{x}\cos{x} \)
    3. \( -\sin{(2x)} \)
    4. \( 2\cos{x} \)
    5. \( -2\sin{x} \)
    12. Solution C
  12. \( y = x^2 \sin{ \left( \frac{1}{x} \right) } \text{, } (x \neq 0 ) \)
    1. \( 2x\sin{ \left( \frac{1}{x} \right)} - x^2\cos{ \left( \frac{1}{x} \right)} \)
    2. \( -\frac{2}{x}\cos{ \left( \frac{1}{x} \right)} \)
    3. \( 2x\cos{ \left( \frac{1}{x} \right)} \)
    4. \( 2x\sin{ \left( \frac{1}{x} \right)} - \cos{ \left( \frac{1}{x} \right)} \)
    5. \( -\cos{ \left( \frac{1}{x} \right)} \)
    13. Solution D
  13. \( \frac{1}{2\sin{(2x)}} \)
    1. \( -\csc{(2x)}\cot{(2x)} \)
    2. \( \frac{1}{4\cos{(2x)}} \)
    3. \( -4\csc{(2x)}\cot{(2x)} \)
    4. \( \frac{\cos{(2x)}}{2\sqrt{\sin{(2x)}}} \)
    5. \( -\csc^2{(2x)} \)
    14. Solution A
  14. \( y = \sec^2{\sqrt{x}} \)
    1. \( \frac{\sec{\sqrt{x}}\tan{\sqrt{x}}}{{\sqrt{x}}} \)
    2. \( \frac{\tan{\sqrt{x}}}{{\sqrt{x}}} \)
    3. \( 2\sec{\sqrt{x}}\tan^2{\sqrt{x}} \)
    4. \( \frac{\sec^2{\sqrt{x}}\tan{\sqrt{x}}}{{\sqrt{x}}} \)
    5. \( 2\sec^2 { \left( x\sqrt{x} \right) } \tan{\sqrt{x}} \)
    15. Solution D
  15. \( \frac{1 + x^2}{1 - x^2} \)
    1. \( -\frac{4x}{ \left( 1 - x^2 \right) ^2} \)
    2. \( \frac{4x}{ \left( 1 - x^2 \right) ^2} \)
    3. \( -\frac{4x^3}{ \left( 1 - x^2 \right) ^2} \)
    4. \( \frac{2x}{ 1 - x^2 } \)
    5. \( \frac{4}{ 1 - x^2 } \)
    16. Solution B
  16. \( y = \sin^{-1}{x} - \sqrt{1 - x^2} \)
    1. \( \frac{1}{2\sqrt{1 - x^2}} \)
    2. \( \frac{2}{\sqrt{1 - x^2}} \)
    3. \( \frac{1 + x}{\sqrt{1 - x^2}} \)
    4. \( \frac{x^2}{\sqrt{1 - x^2}} \)
    5. \( \frac{1}{\sqrt{1 - x^2}} \)
    17. Solution C
  17. \( x = t\sin{t} \text{ and } y = 1 - \cos{t} \)
    1. \( \frac{\sin{t}}{t \cos{t} + \sin{t}} \)
    2. \( \frac{1 - \cos{t}}{\sin{t}} \)
    3. \( \frac{\sin{t}}{\cos{t} - 1} \)
    4. \( \frac{1 - x}{y} \)
    5. \( \frac{1 - \cos{t}}{1 - \sin{t}} \)
    18. Solution A
  18. \( x = \cos^3 {\theta} \text{ and } y = \sin^3{\theta} \)
    1. \( \tan^3{\theta} \)
    2. \( -cot{\theta} \)
    3. \( cot{\theta} \)
    4. \( -tan{\theta} \)
    5. \( -tan^2{\theta} \)
    19. Solution D
  19. \( x^3 - xy + y^3 = 1 \)
    1. \( \frac{3x^2}{x - 3y^2} \)
    2. \( \frac{3x^2 - 1}{1 - 3y^2} \)
    3. \( \frac{y - 3x^2}{3y^2 - x} \)
    4. \( \frac{3x^2 + 3y^2 - y}{x} \)
    5. \( \frac{3x^2 + 3y^2}{x} \)
    20. Solution C
  20. \( x + \cos{(x + y)} = 0 \)
    1. \( \csc{(x + y)} - 1 \)
    2. \( \csc{(x + y)} \)
    3. \( \frac{x}{\sin{(x + y)}} \)
    4. \( \frac{1}{\sqrt{1 - x^2}} \)
    5. \( \frac{1 - \sin{x}}{\sin{y}} \)
    21. Solution A
  21. \( \sin{x} - \cos{y} - 2 = 0 \)
    1. \( -\cot{x} \)
    2. \( -\cot{y} \)
    3. \( \frac{\cos{x}}{\sin{y}} \)
    4. \( -\csc{y}\cos{x} \)
    5. \( \frac{2 - \cos{x}}{\sin{y}} \)
    22. Solution D
  22. \( 3x^2 - 2xy + 5y^2 = 1 \)
    1. \( \frac{3x + y}{x - 5y} \)
    2. \( \frac{y - 3x}{5y - x} \)
    3. \( 3x + 5y \)
    4. \( \frac{3x + 4y}{x} \)
    5. None of these
    23. Solution B
Answer the following:
  1. If \(x = t^2 - 1\text{ and }y = t^4 - 2t^3,\) then \( \frac{d^2y}{dx^2} \mid_{t = 1} = \)
    1. 1
    2. -1
    3. 0
    4. 3
    5. \( \frac{1}{2} \)
    2. Solution E
  2. If \( f(x) = x^4 - 4x^3 + 4x^2 - 1, \) then the set of values of \(x\) for which \(f'(x) = 0\) is
    1. \( \{1, 2\} \)
    2. \( \{0, -1, -2\} \)
    3. \( \{-1, 2\} \)
    4. \( \{0\} \)
    5. \( \{0, 1, 2\} \)
    3. Solution E
  3. If \( f(x) = 16\sqrt{x} \), then \(f''''(4) = \)
    1. \( -\frac{15}{128} \)
    2. -4
    3. \( -\frac{1}{2} \)
    4. \( 0 \)
    5. \( 6 \)
    4. Solution A
  4. If \(f(x) = \frac{x}{ \left( x - 1 \right) ^2},\) then the set of all values of \(x\) for which \(f'(x)\) exists is:
    1. \( x \in \Re \)
    2. \( x \neq \pm 1 \)
    3. \( x \neq - 1 \)
    4. \( x \neq \frac{1}{3}\text{, }x \neq -1 \)
    5. \( x \neq 1 \)
    5. Solution E
  5. If \( f(x) = \frac{1}{x^2 + 1} \) and \(g(x) = \sqrt{x}\), then \(\frac{d}{dx} f(g(x)) =\)
    1. \( -\frac{\sqrt{x}}{ \left( x^2 + 1 \right) ^2} \)
    2. \( - (x + 1)^{-2} \)
    3. \(-\frac{2x}{\left( x^2 + 1 \right) ^2} \)
    4. \( \frac{1}{\left( x^2 + 1 \right) ^2} \)
    5. \( \frac{1}{2\sqrt{x}(x + 1)} \)
    6. Solution B
  6. If \( x = \cos{t}\) and \(y = \cos{(2t)},\) then \(\frac{d^2y}{dx^2} = \)
    1. \( 4\cos{t} \)
    2. 4
    3. \( \frac{4y}{x} \)
    4. -4
    5. \( -4\cot{t} \)
    7. Solution B. This one is easiest if, when you get \( \frac{dy}{dx} = \frac{-2\sin{(2t)}}{-sint}\) you apply the double angle formula to the numerator and reduce before you differentiate the second time.
  7. Suppose \( y = f(x)\text{ and }x = f^{-1}(y) \) are mutually inverse functions. If \(f(1) = 4\) and \( \frac{dy}{dx} = -3\) at \(x = 1\), then \(\frac{dx}{dy} \mid_{y = 4} = \)
    1. \( -\frac{1}{3} \)
    2. \( -\frac{1}{4} \)
    3. \( \frac{1}{3} \)
    4. 3
    5. 4
    8. Solution A