Calculus/Differentiation/Basics of Differentiation/Solutions

From Wikibooks, open books for an open world
Jump to: navigation, search

Find the Derivative by Definition[edit]

1. f(x) = x^2 \,

\begin{align}f'(x)
&=\lim_{\Delta x \to 0}\frac{(x+\Delta x)^2-x^2}{\Delta x}\\
&=\lim_{\Delta x \to 0}\frac{x^2+2x\Delta x+\Delta x^2-x^2}{\Delta x}\\
&=\lim_{\Delta x \to 0}\frac{2x\Delta x+\Delta x^2}{\Delta x}\\
&=\lim_{\Delta x \to 0}2x+\Delta x\\
&=\mathbf{2x}\end{align}

2. f(x) = 2x + 2 \,

\begin{align}f'(x)
&=\lim_{\Delta x \to 0}\frac{[2(x+\Delta x) + 2] - (2x + 2)}{\Delta x}\\
&=\lim_{\Delta x \to 0}\frac{2x+2\Delta x + 2 - 2x - 2}{\Delta x}\\
&=\lim_{\Delta x \to 0}\frac{2\Delta x}{\Delta x}\\
&=\mathbf{2}\end{align}

3. f(x) = \frac{1}{2}x^2 \,

\begin{align}f'(x)
&=\lim_{\Delta x \to 0}\frac{\frac{1}{2}(x+\Delta x)^2-\frac{1}{2}x^2}{\Delta x}\\
&=\lim_{\Delta x \to 0}\frac{\frac{1}{2}(x^2+2x\Delta x+\Delta x^2)-\frac{1}{2}x^2}{\Delta x}\\
&=\lim_{\Delta x \to 0}\frac{\frac{x^2}{2}+\frac{2x \Delta x}{2}+\frac{\Delta x^2}{2}-\frac{x^2}{2}}{\Delta x}\\
&=\lim_{\Delta x \to 0}\frac{2x\Delta x + \Delta x^2}{2\Delta x}\\
&=\lim_{\Delta x \to 0}x+\frac{\Delta x}{2}\\
&=\mathbf{x}\end{align}

4. f(x) = 2x^2 + 4x + 4 \,

\begin{align}f'(x)
&=\lim_{\Delta x \to 0}\frac{[2(x+\Delta x)^2 + 4(x+\Delta x)+4] - (2x^2+4x+4)}{\Delta x}\\
&=\lim_{\Delta x \to 0}\frac{4x\Delta x + 2\Delta x^2 + 4\Delta x}{\Delta x}\\
&=\lim_{\Delta x \to 0}4x+2\Delta x + 4\\
&=\mathbf{4x + 4}\end{align}

5. f(x) = \sqrt{x+2} \,

\begin{align}f'(x)
&=\lim_{\Delta x \to 0}\frac{\sqrt{x+\Delta x+2}-\sqrt{x+2}}{\Delta x}\\
&=\lim_{\Delta x \to 0}(\frac{\sqrt{x+\Delta x+2}-\sqrt{x+2}}{\Delta x})(\frac{\sqrt{x+\Delta x+2}+\sqrt{x+2}}{\sqrt{x+\Delta x+2}+\sqrt{x+2}})\\
&=\lim_{\Delta x \to 0}\frac{x+\Delta x+2-x-2}{\Delta x(\sqrt{x+\Delta x+2}+\sqrt{x+2})}\\
&=\lim_{\Delta x \to 0}\frac{\Delta x}{\Delta x(\sqrt{x+\Delta x+2}+\sqrt{x+2})}\\
&=\lim_{\Delta x \to 0}\frac{1}{\sqrt{x+\Delta x+2}+\sqrt{x+2}}\\
&=\mathbf{\frac{1}{2\sqrt{x+2}}}\end{align}

6. f(x) = \frac{1}{x} \,

\begin{align}f'(x)
&=\lim_{\Delta x \to 0}\frac{\frac{1}{x+\Delta x}-\frac{1}{x}}{\Delta x}\\
&=\lim_{\Delta x \to 0}\frac{\frac{x-x-\Delta x}{x(x+\Delta x)}}{\Delta x}\\
&=\lim_{\Delta x \to 0}\frac{-\Delta x}{x\Delta x(x+\Delta x)}\\
&=\lim_{\Delta x \to 0}\frac{-1}{x(x+\Delta x)}\\
&=\mathbf{-\frac{1}{x^2}}\end{align}

7. f(x) = \frac{3}{x+1} \,

\begin{align}f'(x)
&=\lim_{\Delta x \to 0}\frac{\frac{3}{x+\Delta x+1}-\frac{3}{x+1}}{\Delta x}\\
&=\lim_{\Delta x \to 0}\frac{\frac{3x+3-(3x+3\Delta x+3)}{(x+1)(x+\Delta x+1)}}{\Delta x}\\
&=\lim_{\Delta x \to 0}\frac{-3\Delta x}{\Delta x(x+1)(x+\Delta x+1)}\\
&=\lim_{\Delta x \to 0}\frac{-3}{(x+1)(x+\Delta x+1)}\\
&=\mathbf{\frac{-3}{(x+1)^2}}\end{align}

8. f(x) = \frac{1}{\sqrt{x+1}} \,

\begin{align}f'(x)
&=\lim_{\Delta x \to 0}\frac{\frac{1}{\sqrt{x+\Delta x+1}}-\frac{1}{\sqrt{x+1}}}{\Delta x}\\
&=\lim_{\Delta x \to 0}\frac{\frac{\sqrt{x+1}-\sqrt{x+\Delta x+1}}{\sqrt{x+\Delta x+1}\sqrt{x+1}}}{\Delta x}\\
&=\lim_{\Delta x \to 0}(\frac{\sqrt{x+1}-\sqrt{x+\Delta x+1}}{\Delta x\sqrt{x+\Delta x+1}\sqrt{x+1}})(\frac{\sqrt{x+1}+\sqrt{x+\Delta x+1}}{\sqrt{x+1}+\sqrt{x+\Delta x+1}})\\
&=\lim_{\Delta x \to 0}\frac{x+1-(x+\Delta x+1)}{\Delta x\sqrt{x+\Delta x+1}\sqrt{x+1}(\sqrt{x+\Delta x+1}+\sqrt{x+1})}\\
&=\lim_{\Delta x \to 0}\frac{-1}{\sqrt{x+\Delta x+1}\sqrt{x+1}(\sqrt{x+\Delta x+1}+\sqrt{x+1})}\\
&=\frac{-1}{(x+1)(2\sqrt{x+1})}\\
&=\mathbf{\frac{-1}{2(x+1)^{3/2}}}\end{align}

9. f(x) = \frac{x}{x+2} \,

\begin{align}f'(x)
&=\lim_{\Delta x \to 0}\frac{\frac{x+\Delta x}{x+\Delta x+2}-\frac{x}{x+2}}{\Delta x}\\
&=\lim_{\Delta x \to 0}\frac{(x+\Delta x)(x+2)-x(x+\Delta x+2)}{\Delta x(x+\Delta x+2)(x+2)}\\
&=\lim_{\Delta x \to 0}\frac{x^2+2x+x\Delta x+2\Delta x-x^2-x\Delta x-2x}{\Delta x(x+\Delta x+2)(x+2)}\\
&=\lim_{\Delta x \to 0}\frac{2\Delta x}{\Delta x(x+\Delta x+2)(x+2)}\\
&=\lim_{\Delta x \to 0}\frac{2}{(x+\Delta x+2)(x+2)}\\
&=\mathbf{\frac{2}{(x+2)^2}}\end{align}

Prove the Constant Rule[edit]

10. Use the definition of the derivative to prove that for any fixed real number c, \frac{d}{dx}\left[cf(x)\right] = c \frac{d}{dx}\left[f(x)\right]

\begin{align}\frac{d}{dx}\left[cf(x)\right]
&=\lim_{\Delta x \to 0}\frac{cf\left(x+\Delta x \right)-cf\left(x\right)}{\Delta x}\\
&=c\lim_{\Delta x \to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}\\
&=c\frac{d}{dx}\left[f(x)\right]\end{align}

Find the Derivative by Rules[edit]

Power Rule[edit]

11. f(x) = 2x^2 + 4\,

f'(x)=\mathbf{4x}

12. f(x) = 3\sqrt[3]{x}\,

f'(x)=3(\frac{1}{3})x^{-2/3}=\mathbf{\frac{1}{\sqrt[3]{x^2}}}

13. f(x) = 2x^5+8x^2+x-78\,

f'(x)=\mathbf{10x^4+16x+1}

14. f(x) = 7x^7+8x^5+x^3+x^2-x\,

f'(x)=\mathbf{49x^6+40x^4+3x^2+2x-1}

15. f(x) = \frac{1}{x^2}+3x^\frac{1}{3}\,

f'(x)=\frac{-2}{x^{3}}+x^{-2/3}=\mathbf{\frac{-2}{x^3}+\frac{1}{\sqrt[3]{x^2}}}

16. f(x) = 3x^{15} + \frac{1}{17}x^2 +\frac{2}{\sqrt{x}} \,

f'(x)=45x^{14}+\frac{2}{17}x-\frac{1}{\sqrt{x^{3}}}=\mathbf{45x^{14}+\frac{2}{17}x-\frac{1}{x\sqrt{x}}}

17. f(x) = \frac{3}{x^4} - \sqrt[4]{x} + x \,

f'(x)=\frac{-12}{x^{5}}-\frac{1}{4}x^{-3/4}+1=\mathbf{\frac{-12}{x^5}-\frac{1}{4\sqrt[4]{x^3}}+1}

18. f(x) = 6x^{1/3}-x^{0.4} +\frac{9}{x^2} \,

f'(x)=2x^{-2/3}-0.4x^{-0.6}-\frac{18}{x^{3}}=\mathbf{\frac{2}{\sqrt[3]{x^2}}-\frac{0.4}{x^{0.6}}-\frac{18}{x^3}}

19. f(x) = \frac{1}{\sqrt[3]{x}} + \sqrt{x} \,

f'(x)=-\frac{1}{3x^{4/3}}+\frac{1}{2\sqrt{x}}=\mathbf{\frac{-1}{3x\sqrt[3]{x}}+\frac{1}{2\sqrt{x}}}

Product Rule[edit]

20. f(x) = (x^4+4x+2)(2x+3) \,

f'(x)=(4x^{3}+4)(2x+3)+(x^{4}+4x+2)(2)=\mathbf{10x^4+12x^3+16x+16}

21. f(x) = (2x-1)(3x^2+2) \,

f'(x)=(2)(3x^{2}+2)+(2x-1)(6x)=\mathbf{18x^2-6x+4}

22. f(x) = (x^3-12x)(3x^2+2x) \,

f'(x)=(3x^{2}-12)(3x^{2}+2x)+(x^{3}-12x)(6x+2)=\mathbf{15x^4+8x^3-108x^2-48x}

23. f(x) = (2x^5-x)(3x+1) \,

f'(x)=(10x^{4}-1)(3x+1)+(2x^{5}-x)(3)=\mathbf{36x^5+10x^4-6x-1}

Quotient Rule[edit]

24. f(x) = \frac{2x+1}{x+5} \,

f'(x)=\frac{(x+5)(2)-(2x+1)}{(x+5)^{2}}=\mathbf{\frac{9}{(x+5)^2}}

25. f(x) = \frac{3x^4+2x +2}{3x^2+1} \,

f'(x)=\frac{(3x^{2}+1)(12x^{3}+2)-(3x^{4}+2x+2)(6x)}{(3x^{2}+1)^{2}}=\mathbf{\frac{18x^5+12x^3-6x^2-12x+2}{(3x^2+1)^2}}

26. f(x) = \frac{x^\frac{3}{2}+1}{x+2} \,

f'(x)=\frac{(x+2)(\frac{3}{2}\sqrt{x})-(x^{\frac{3}{2}}+1)}{(x+2)^2}=\mathbf{\frac{x\sqrt{x}+6\sqrt{x}-2}{2(x+2)^2}}

27. d(u) = \frac{u^3+2}{u^3} \,

d'(u)=\frac{u^{3}(3u^{2})-(u^{3}+2)(3u^{2})}{u^{6}}=\mathbf{-\frac{6}{u^4}}

28. f(x) = \frac{x^2+x}{2x-1} \,

f'(x)=\frac{(2x-1)(2x+1)-(x^{2}+x)(2)}{(2x-1)^{2}}=\mathbf{\frac{2x^2-2x-1}{(2x-1)^2}}

29. f(x) = \frac{x+1}{2x^2+2x+3} \,

f'(x)=\frac{(2x^{2}+2x+3)-(x+1)(4x+2)}{(2x^{2}+2x+3)^2}=\mathbf{\frac{-2x^2-4x+1}{(2x^2+2x+3)^2}}

30. f(x) = \frac{16x^4+2x^2}{x} \,

f'(x)=\frac{x(64x^{3}+4x)-(16x^{4}+2x^{2})}{x^{2}}=\mathbf{48x^2+2}

Chain Rule[edit]

31. f(x) = (x+5)^2 \,

Let f(x)=g(h(x));\quad g(x)=x^{2};\quad h(x)=x+5. Then
f'(x)=\frac{dg}{dh}\frac{dh}{dx}=2(x+5)(1)=\mathbf{2(x+5)}

32. g(x) = (x^3 - 2x + 5)^2 \,

Let g(x)=r(s(x));\quad r(x)=x^{2};\quad s(x)=x^{3}-2x+5. Then
g'(x)=\frac{dr}{ds}\frac{ds}{dx}=\mathbf{2(x^{3}-2x+5)(3x^{2}-2)}

33. f(x) = \sqrt{1-x^2} \,

Let f(x)=g(h(x));\quad g(x)=\sqrt{x};\quad h(x)=1-x^{2}. Then
f'(x)=\frac{dg}{dh}\frac{dh}{dx}=\frac{1}{2\sqrt{1-x^{2}}}(-2x)=\mathbf{-\frac{x}{\sqrt{1-x^{2}}}}

34. f(x) = \frac{(2x+4)^3}{4x^3+1} \,

Let f(x)=\frac{N(x)}{D(x)};\quad N(x)=g(h(x));\quad g(x)=x^{3};\quad h(x)=2x+4;\quad D(x)=4x^{3}+1. Then
f'(x)=\frac{D(x)N'(x)-N(x)D'(x)}{D^{2}(x)}
N'(x)=\frac{dg}{dh}\frac{dh}{dx}=3(2x+4)^{2}(2)=6(2x+4)^{2}
D'(x)=12x^{2}
f'(x)=\frac{(4x^{3}+1)6(2x+4)^{2}-(2x+4)^{3}12x^{2}}{(4x^{3}+1)^{2}}=\mathbf{\frac{6(4x^{3}+1)(2x+4)^{2}-12x^2(2x+4)^{3}}{(4x^{3}+1)^{2}}}

35. f(x) = (2x+1)\sqrt{2x+2} \,

Let f(x)=A(x)B(x);\quad A(x)=2x+1;\quad B(x)=g(h(x));\quad g(x)=\sqrt{x};\quad h(x)=2x+2. Then
f'(x)=A'(x)B(x)+A(x)B'(x)
A'(x)=2
B'(x)=\frac{dg}{dh}\frac{dh}{dx}=\frac{1}{2\sqrt{2x+2}}(2)=\frac{1}{\sqrt{2x+2}}
f'(x)=\mathbf{2\sqrt{2x+2}+\frac{2x+1}{\sqrt{2x+2}}}

36. f(x) = \frac{2x+1}{\sqrt{2x+2}} \,

Let f(x)=\frac{N(x)}{D(x)};\quad N(x)=2x+1;\quad D(x)=g(h(x));\quad g(x)=\sqrt{x};\quad h(x)=2x+2. Then
f'(x)=\frac{D(x)N'(x)-N(x)D'(x)}{D^{2}(x)}
N'(x)=2
D'(x)=\frac{dg}{dh}\frac{dh}{dx}=\frac{1}{2\sqrt{2x+2}}(2)=\frac{1}{\sqrt{2x+2}}
f'(x)=\frac{\sqrt{2x+2}(2)-\frac{(2x+1)}{\sqrt{2x+2}}}{2x+2}=\mathbf{\frac{2x+3}{(2x+2)^{3/2}}}

37. f(x) = \sqrt{2x^2+1}(3x^4+2x)^2 \,

Let f(x)=A(x)B(x);\quad A(x)=g(h(x));\quad g(x)=\sqrt{x};\quad h(x)=2x^{2}+1;\quad B(x)=r(s(x));\quad r(x)=x^{2};\quad s(x)=3x^{4}+2x. Then
f'(x)=A'(x)B(x)+A(x)B'(x)
A'(x)=\frac{dg}{dh}\frac{dh}{dx}=\frac{1}{2\sqrt{2x^{2}+1}}(4x)=\frac{2x}{\sqrt{2x^{2}+1}}
B'(x)=\frac{dr}{ds}\frac{ds}{dx}=2(3x^{4}+2x)(12x^{3}+2)
f'(x)=\frac{2x}{\sqrt{2x^{2}+1}}(3x^{4}+2x)^{2}+\sqrt{2x^{2}+1}(2)(3x^{4}+2x)(12x^{3}+2)=\mathbf{\frac{2x(3x^{4}+2x)^{2}}{\sqrt{2x^{2}+1}}+\sqrt{2x^{2}+1}(2)(3x^{4}+2x)(12x^{3}+2)}

38. f(x) = \frac{2x+3}{(x^4+4x+2)^2} \,

Let f(x)=\frac{N(x)}{D(x)};\quad N(x)=2x+3;\quad D(x)=g(h(x));\quad g(x)=x^{2};\quad h(x)=x^{4}+4x+2. Then
f'(x)=\frac{D(x)N'(x)-N(x)D'(x)}{D^{2}(x)}
N'(x)=2
D'(x)=\frac{dg}{dh}\frac{dh}{dx}=2(x^{4}+4x+2)(4x^{3}+4)
f'(x)=\frac{(x^{4}+4x+2)^{2}(2)-(2x+3)(2)(x^{4}+4x+2)(4x^{3}+4)}{(x^{4}+4x+2)^{4}}=\mathbf{\frac{2(x^{4}+4x+2)^{2}-2(2x+3)(x^{4}+4x+2)(4x^{3}+4)}{(x^{4}+4x+2)^{4}}}

39. f(x) = \sqrt{x^3+1}(x^2-1) \,

Let f(x)=A(x)B(x);\quad A(x)=g(h(x));\quad g(x)=\sqrt{x};\quad h(x)=x^{3}+1;\quad B(x)=x^{2}-1. Then
f'(x)=A'(x)B(x)+A(x)B'(x)
A'(x)=\frac{dg}{dh}\frac{dh}{dx}=\frac{1}{2\sqrt{x^{3}+1}}(3x)=\frac{3x}{2\sqrt{x^{3}+1}}
B'(x)=2x
f'(x)=\frac{3x}{2\sqrt{x^{3}+1}}(x^{2}-1)+\sqrt{x^{3}+1}(2x)=\mathbf{\frac{3x(x^{2}-1)}{2\sqrt{x^{3}+1}}+2x\sqrt{x^{3}+1}}

40. f(x) = ((2x+3)^4 + 4(2x+3) +2)^2 \,

Let f(x)=g((h(x));\quad g(x)=x^{2};\quad h(x)=A(x)+B(x)+2;\quad A(x)=r(s(x));\quad r(x)=x^{4};\quad s(x)=2x+3;\quad B(x)=4(2x+3). Then
f'(x)=\frac{dg}{dh}\frac{dh}{dx}=2(A(x)+B(x)+2)(A'(x)+B'(x))
A'(x)=\frac{dr}{ds}\frac{ds}{dx}=4(2x+3)^{3}(2)=8(2x+3)^{3}
B'(x)=8
f'(x)=\mathbf{2((2x+3)^{4}+4(2x+3)+2)(8(2x+3)^{3}+8)}

41. f(x) = \sqrt{1+x^2} \,

Let f(x)=g(h(x));\quad g(x)=\sqrt{x};\quad h(x)=1+x^{2}. Then
f'(x)=\frac{dg}{dh}\frac{dh}{dx}=\frac{1}{2\sqrt{1+x^{2}}}(2x)=\mathbf{\frac{x}{\sqrt{1+x^{2}}}}

Exponentials[edit]

42. f(x) = (3x^2+e)e^{2x}\,

f'(x)=f'(x)=(6x)e^{2x}+(3x^{2}+e)2e^{2x}=\mathbf{6xe^{2x}+2e^{2x}(3x^{2}+e)}

43. f(x) = e^{2x^2+3x}

Let f(x)=g(h(x));\quad g(x)=e^{x};\quad h(x)=2x^{2}+3x. Then
f'(x)=\frac{dg}{dh}\frac{dh}{dx}=e^{2x^{2}+3x}(4x+3)=\mathbf{(4x+3)e^{2x^{2}+3x}}

44. f(x) = e^{e^{2x^2+1}}

Let

u(x)=e^{x}
v(x)=e^{x}
w(x)=2x^{2}+1

Then

f(x)=u(v(w(x)))

Using the chain rule, we have

\frac{df}{dx}=\frac{du}{dv}\frac{dv}{dw}\frac{dw}{dx}

The individual factor are

\frac{du}{dv}=\frac{d(e^{v})}{dv}=e^{v}=e^{e^{w}}=e^{e^{2x^{2}+1}}
\frac{dv}{dw}=\frac{d(e^{w})}{dw}=e^{w}=e^{2x^{2}+1}
\frac{dw}{dx}=\frac{d(2x^{2}+1)}{dx}=4x

So

\frac{df}{dx}=(e^{e^{2x^{2}+1}})(e^{2x^{2}+1})(4x)=\mathbf{4xe^{2x^{2}+1+e^{2x^{2}+1}}}
45. f(x) = 4^x\,

f'(x)=\mathbf{\ln(4)4^{x}}

Logarithms[edit]

46. f(x) = 2^{x-3}\cdot3\sqrt{x^3-2}+\ln x\,

f'(x)=\ln(2)2^{x-3}\cdot3\sqrt{x^{3}-2}+2^{x-3}\frac{3}{2\sqrt{x^{3}-2}}3x^{2}+\frac{1}{x}=\mathbf{3\ln(2)2^{x-3}\sqrt{x^{3}-2}+\frac{9x^{2}2^{x-4}}{\sqrt{x^{3}-2}}+\frac{1}{x}}

47. f(x) = \ln x - 2e^x + \sqrt{x}\,

f'(x)=\mathbf{\frac{1}{x}-2e^{x}+\frac{1}{2\sqrt{x}}}

48. f(x) = \ln(\ln(x^3(x+1))) \,

Let f(x)=g(g(h(x)));\quad g(x)=\ln(x);\quad h(x)=x^{3}(x+1). Then
f'(x)=\frac{dg(g(h(x)))}{dg(h(x))}\frac{dg(h(x))}{dh(x)}\frac{dh(x)}{dx}=\frac{1}{\ln(x^{3}(x+1))}\frac{1}{x^{3}(x+1)}(4x^{3}+3x^{2})=\mathbf{\frac{4x^{3}+3x^{2}}{x^{3}(x+1)\ln(x^{3}(x+1))}}

49. f(x) = \ln(2x^2+3x)\,

f'(x)=\frac{1}{2x^{2}+3x}(4x+3)=\mathbf{\frac{4x+3}{2x^{2}+3x}}

50. f(x) = \log_4 x + 2\ln x\,

f'(x)=\mathbf{\frac{1}{x\ln4}+\frac{2}{x}}

Trigonometric functions[edit]

51. f(x) = 3e^x-4\cos (x) - \frac{1}{4}\ln x\,

f'(x)=\mathbf{3e^{x}+4\sin(x)-\frac{1}{4x}}

52. f(x) = \sin(x)+\cos(x)\,

f'(x)=\mathbf{\cos(x)-\sin(x)}

More Differentiation[edit]

53. \frac{d}{dx}[(x^{3}+5)^{10}]

10(x^{3}+5)^{9}(3x^{2})=\mathbf{30x^{2}(x^{3}+5)^{9}}

54. \frac{d}{dx}[x^{3}+3x]

\mathbf{3x^{2}+3}

55. \frac{d}{dx}[(x+4)(x+2)(x-3)]

Let f(x)=A(x)B(x)C(x);\quad A(x)=x+4;\quad B(x)=x+2;\quad C(x)=x-3. Then
f'(x)=A'(x)B(x)C(x)+A(x)B'(x)C(x)+A(x)B(x)C'(x)
A'(x)=B'(x)=C'(x)=1
f'(x)=\mathbf{(x+2)(x-3)+(x+4)(x-3)+(x+4)(x+2)}

56. \frac{d}{dx}[\frac{x+1}{3x^{2}}]

\frac{3x^{2}-(x+1)(6x)}{(3x^{2})^{2}}=\frac{3x^{2}-6x^{2}-6x}{9x^{4}}=\frac{-3x^{2}-6x}{9x^{4}}=\mathbf{-\frac{x+2}{3x^{3}}}

57. \frac{d}{dx}[3x^{3}]

\mathbf{9x^{2}}

58. \frac{d}{dx}[x^{4}\sin x]

\mathbf{4x^{3}\sin x+x^{4}\cos x}

59. 2^{x}

\mathbf{\ln(2)2^{x}}

60. \frac{d}{dx}[e^{x^{2}}]

\mathbf{2xe^{x^{2}}}

61. \frac{d}{dx}[e^{2^{x}}]

\mathbf{\ln(2)2^{x}e^{2^{x}}}

Implicit Differentiation[edit]

Use implicit differentiation to find y'

62.  x^3 + y^3 = xy \,

3x^{2}+3y^{2}y'=y+xy'
3y^{2}y'-xy'=y-3x^{2}
\mathbf{y'=\frac{y-3x^{2}}{3y^{2}-x}}

63.  (2x+y)^4 + 3x^2 +3y^2 = \frac{x}{y} + 1 \,

4(2x+y)^{3}(2+y')+6x+6yy'=\frac{y-xy'}{y^{2}}
8(2x+y)^{3}+4y'(2x+y)^{3}+6x+6yy'=\frac{1}{y}-\frac{xy'}{y^{2}}
y'(4(2x+y)^{3}+6y+\frac{x}{y^{2}})=\frac{1}{y}-8(2x+y)^{3}-6x
y'(\frac{4y^{2}(2x+y)^{3}+6y^{3}+x}{y^{2}})=\frac{y-8y^{2}(2x+y)^{3}-6xy^{2}}{y^{2}}
\mathbf{y'=\frac{y-8y^{2}(2x+y)^{3}-6xy^{2}}{4y^{2}(2x+y)^{3}+6y^{3}+x}}

Logarithmic Differentiation[edit]

Use logarithmic differentiation to find \frac{dy}{dx}:

64. y = x(\sqrt[4]{1-x^3}\,)

\ln y=\ln(x)+\ln(\sqrt[4]{1-x^{3}})=\ln(x)+\frac{\ln(1-x^{3})}{4}
\frac{y'}{y}=\frac{1}{x}-\frac{3x^{2}}{4(1-x^{3})}
y'=x(\sqrt[4]{1-x^{3}}\,)(\frac{1}{x}-\frac{3x^{2}}{4(1-x^{3})})=\mathbf{\sqrt[4]{1-x^{3}}-\frac{3x^{3}}{4(1-x^{3})^{3/4}}}

65. y = \sqrt{x+1 \over 1-x}\,

\ln y=\frac{1}{2}(\ln(x+1)-\ln(1-x))
\frac{y'}{y}=\frac{1}{2}(\frac{1}{x+1}+\frac{1}{1-x})
\mathbf{y'=\frac{1}{2}\sqrt{\frac{x+1}{1-x}}(\frac{1}{x+1}+\frac{1}{1-x})}

66. y = (2x)^{2x}\,

\ln y=2x\ln(2x)
\frac{y'}{y}=2\ln(2x)+2x\frac{2}{2x}=2\ln(2x)+2
\mathbf{y'=(2x)^{2x}(2\ln(2x)+2)}

67. y = (x^3+4x)^{3x+1}\,

\ln y=(3x+1)\ln(x^{3}+4x)
\frac{y'}{y}=3\ln(x^{3}+4x)+(3x+1)\frac{3x^{2}+4}{x^{3}+4x}
\mathbf{y'=(x^{3}+4x)^{3x+1}(3\ln(x^{3}+4x)+\frac{(3x+1)(3x^{2}+4)}{x^{3}+4x})}

68. y = (6x)^{\cos(x) + 1}\,

\ln y=(\cos(x)+1)\ln(6x)
\frac{y'}{y}=-\sin(x)\ln(6x)+\frac{\cos(x)+1}{x}
\mathbf{y'=(6x)^{\cos(x)+1}(-\sin(x)\ln(6x)+\frac{\cos(x)+1}{x})}

Equation of Tangent Line[edit]

For each function, f, (a) determine for what values of x the tangent line to f is horizontal and (b) find an equation of the tangent line to f at the given point.

69.  f(x) = \frac{x^3}{3} + x^2 + 5, \;\;\; (3,23)
f'(x)=x^{2}+2x

a) x^{2}+2x=0\implies \mathbf{x=0,-2}
b) m=3^{2}+2(3)=9+6=15

23=15(3)+b\implies b=-22
\mathbf{y=15x-22}
70.  f(x) = x^3 - 3x + 1, \;\;\;  (1,-1)
f'(x)=3x^{2}-3

a) 3x^{2}-3=0\implies \mathbf{x=\pm1}
b) m=3(1)^{2}-3=0

-1=b
\mathbf{y=-1}
71.  f(x) = \frac{2}{3} x^3 + x^2 - 12x + 6, \;\;\; (0,6)
f'(x)=2x^{2}+2x-12

a) 2x^{2}+2x-12=0=x^{2}+x-6=(x+3)(x-2)\implies \mathbf{x=2,-3}
b) m=-12

6=b
\mathbf{y=-12x+6}
72.  f(x) = 2x + \frac{1}{\sqrt{x}}, \;\;\; (1,3)
f'(x)=2-\frac{1}{2x^{3/2}}

a) 2-\frac{1}{2x^{3/2}}=0\implies2x^{3/2}=\frac{1}{2}\implies x^{3/2}=\frac{1}{4}=2^{-2}\implies \mathbf{x=2^{-4/3}}
b) m=2-\frac{1}{2(1)^{3/2}}=2-\frac{1}{2}=\frac{3}{2}

3=\frac{3}{2}(1)+b\implies b=\frac{3}{2}
\mathbf{y=\frac{3}{2}x+\frac{3}{2}}
73.  f(x) = (x^2+1)(2-x), \;\;\; (2,0)
f'(x)=(2x)(2-x)-(x^{2}+1)=4x-2x^{2}-x^{2}-1=-3x^{2}+4x-1

a) -3x^{2}+4x-1=0=(3x-1)(-x+1)\implies \mathbf{x=1,\frac{1}{3}}
b) m=-3(2)^{2}+4(2)-1=-3(4)+8-1=-12+7=-5

0=-5(2)+b\implies b=10
\mathbf{y=-5x+10}
74.  f(x) = \frac{2}{3}x^3+\frac{5}{2}x^2 +2x+1, \;\;\; (3,\frac{95}{2})
f'(x)=2x^{2}+5x+2

a) 2x^{2}+5x+2=0=(2x+1)(x+2)\implies \mathbf{x=-\frac{1}{2},-2}
/ b) m=2(3)^{2}+5(3)+2=18+15+2=35

\frac{95}{2}=35(3)+b\implies b=-\frac{115}{2}
\mathbf{y=35x-\frac{115}{2}}
75. Find an equation of the tangent line to the graph defined by (x-y-1)^3 = x \, at the point (1,-1).

3(x-y-1)^{2}(1-y')=1
1-y'=\frac{1}{3(x-y-1)^{2}}
y'=1-\frac{1}{3(x-y-1)^{2}}
m=1-\frac{1}{3(1-(-1)-1)^{2}}=1-\frac{1}{3(1)^{2}}=\frac{2}{3}
-1=\frac{2}{3}(1)+b\implies b=-\frac{5}{3}
\mathbf{y=\frac{2}{3}x-\frac{5}{3}}

76. Find an equation of the tangent line to the graph defined by  e^{xy} + x^2 = y^2 \, at the point (1,0).

y'e^{xy}+2x=2yy'
y'(e^{xy}-2y)=-2x
y'=\frac{2x}{2y-e^{xy}}
m=\frac{2(1)}{2(0)-e^{1(0)}}=\frac{2}{-1}=-2
0=-2(1)+b\implies b=2
\mathbf{y=-2x+2}

Higher Order Derivatives[edit]

77. What is the second derivative of 3x^4+3x^2+2x?

\frac{d}{dx}(3x^4+3x^2+2x)=12x^3+6x+2
\frac{d^2}{dx^2}(3x^4+3x^2+2x)=\frac{d}{dx}(12x^3+6x+2)=\mathbf{36x^2+6}

78. Use induction to prove that the (n+1)th derivative of a n-th order polynomial is 0.

base case: Consider the zeroth-order polynomial, c. \frac{dc}{dx}=0
induction step: Suppose that the n-th derivative of a (n-1)th order polynomial is 0. Consider the n-th order polynomial, f(x). We can write f(x)=cx^n+P(x) where P(x) is a (n-1)th polynomial.
\frac{d^{n+1}}{dx^{n+1}}f(x)=\frac{d^{n+1}}{dx^{n+1}}(cx^n+P(x))=\frac{d^{n+1}}{dx^{n+1}}(cx^n)+\frac{d^{n+1}}{dx^{n+1}}P(x)=\frac{d^{n}}{dx^{n}}(cnx^{n-1})+\frac{d}{dx}\frac{d^{n}}{dx^{n}}P(x)=0+\frac{d}{dx}0=0