# BC Calculus Assignment 18: Differentiation Mechanics

• Class: 5H: BC Calculus
• Author: Peter Atlas
• Text: Calculus Finney, Demana, Waits, Kennedy

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
3. $$y = \frac{2-x}{3x + 1}$$
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. $$\frac{7 - 6x}{(3x + 1)^2}$$
4. Solution A
5. $$y = \sqrt{3 - 2x}$$
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. $$\frac{2}{3} \left( 3 - 2x \right) ^{\frac{3}{2}}$$
6. Solution B
7. $$y=\frac{2}{(5x + 1)^3}$$
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. $$\frac{30}{(5x + 1)^4}$$
8. Solution B
9. $$y= 3x^{\frac{2}{3}} - 4x^{\frac{1}{2}} - 2$$
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. $$2x^{-\frac{1}{3}} - 2x^{-\frac{1}{2}}$$
10. Solution E
11. $$y = 2\sqrt{x} - \frac{1}{2\sqrt{x}}$$
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. $$\frac{4}{\sqrt{x}} + \frac{1}{x\sqrt{x}}$$
12. Solution D
13. $$y = \sqrt{x^2 + 2x - 1}$$
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
14. Solution A
15. $$y=\frac{x}{\sqrt{1 - x^2}}$$
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
16. Solution E
17. $$\cos {\left( x^2\right) }$$
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. $$\sin{(2x)}$$
18. Solution D
19. $$y = \sin^2{(3x)} + \cos^2{(3x)}$$
1. $$-6sin{(6x)}$$
2. $$0$$
3. $$12\sin{(3x)}\cos{(3x)}$$
4. $$6(\sin{(3x)}\cos{(3x)})$$
5. $$1$$
20. Solution B. This is the function $$y = 1$$, whose derivative is everywhere 0.
21. $$y = \cos^2{x}$$
1. $$-\sin^2{x}$$
2. $$2\sin{x}\cos{x}$$
3. $$-\sin{(2x)}$$
4. $$2\cos{x}$$
5. $$-2\sin{x}$$
22. Solution C
23. $$y = x^2 \sin{ \left( \frac{1}{x} \right) } \text{, } (x \neq 0 )$$
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. $$-\cos{ \left( \frac{1}{x} \right)}$$
24. Solution D
25. $$\frac{1}{2\sin{(2x)}}$$
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. $$-\csc^2{(2x)}$$
26. Solution A
27. $$y = \sec^2{\sqrt{x}}$$
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. $$2\sec^2 { \left( x\sqrt{x} \right) } \tan{\sqrt{x}}$$
28. Solution D
29. $$\frac{1 + x^2}{1 - x^2}$$
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. $$\frac{4}{ 1 - x^2 }$$
30. Solution B
31. $$y = \sin^{-1}{x} - \sqrt{1 - x^2}$$
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. $$\frac{1}{\sqrt{1 - x^2}}$$
32. Solution C
33. $$x = t\sin{t} \text{ and } y = 1 - \cos{t}$$
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. $$\frac{1 - \cos{t}}{1 - \sin{t}}$$
34. Solution A
35. $$x = \cos^3 {\theta} \text{ and } y = \sin^3{\theta}$$
1. $$\tan^3{\theta}$$
2. $$-cot{\theta}$$
3. $$cot{\theta}$$
4. $$-tan{\theta}$$
5. $$-tan^2{\theta}$$
36. Solution D
37. $$x^3 - xy + y^3 = 1$$
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. $$\frac{3x^2 + 3y^2}{x}$$
38. Solution C
39. $$x + \cos{(x + y)} = 0$$
1. $$\csc{(x + y)} - 1$$
2. $$\csc{(x + y)}$$
3. $$\frac{x}{\sin{(x + y)}}$$
4. $$\frac{1}{\sqrt{1 - x^2}}$$
5. $$\frac{1 - \sin{x}}{\sin{y}}$$
40. Solution A
41. $$\sin{x} - \cos{y} - 2 = 0$$
1. $$-\cot{x}$$
2. $$-\cot{y}$$
3. $$\frac{\cos{x}}{\sin{y}}$$
4. $$-\csc{y}\cos{x}$$
5. $$\frac{2 - \cos{x}}{\sin{y}}$$
42. Solution D
43. $$3x^2 - 2xy + 5y^2 = 1$$
1. $$\frac{3x + y}{x - 5y}$$
2. $$\frac{y - 3x}{5y - x}$$
3. $$3x + 5y$$
4. $$\frac{3x + 4y}{x}$$
5. None of these
44. Solution B
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
3. 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. $$\{0, 1, 2\}$$
4. Solution E
5. If $$f(x) = 16\sqrt{x}$$, then $$f''''(4) =$$
1. $$-\frac{15}{128}$$
2. -4
3. $$-\frac{1}{2}$$
4. $$0$$
5. $$6$$
6. Solution A
7. 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. $$x \neq 1$$
8. Solution E
9. 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. $$\frac{1}{2\sqrt{x}(x + 1)}$$
10. Solution B
11. 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. $$-4\cot{t}$$
12. 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.
13. 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
14. Solution A