2024/G2/12 - Revision history

Aroldog69360@students.laalliance.org at 16:45, 14 April 2023

2023-04-14T16:45:12Z

← Older revision		Revision as of 16:45, 14 April 2023
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	9) <math> {\frac{d}{dx}} [{\frac{1}{x}}] = {\frac{d}{dx}} [x^{-1}] = -1 \cdot x^{-1-1} = -x^{-2} = - {\frac{1}{x^2}} </math><br><br><br>		9) <math> {\frac{d}{dx}} [{\frac{1}{x}}] = {\frac{d}{dx}} [x^{-1}] = -1 \cdot x^{-1-1} = -x^{-2} = - {\frac{1}{x^2}} </math><br><br><br>

	<math>\mathbf{\color{Purple}{~~Practice Problems~~}}</math><br>		<math>\mathbf{\color{Purple}{PracticeProblems}}</math><br>

	1) <math> {\frac{d}{dx}} [\pi + 3x -5x^{3}] = 0+3-15x^{2} </math><br><br>		1) <math> {\frac{d}{dx}} [\pi + 3x -5x^{3}] = 0+3-15x^{2} </math><br><br>
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	4) <math> {\frac{d}{dx}} [x\sqrt x] = {\frac{d}{dx}} [x^{1} \cdot x^{{\frac{1}{2}}}] = {\frac{d}{dx}} [x^{{\frac{3}{2}}}] = {\frac{3}{2}} \cdot x^{{\frac{3}{2}} -1} = {\frac{3}{2x}}^{{\frac{3}{2}}} - {\frac{2}{2}} = {\frac{3}{2x}}^{{\frac{1}{2}}} = {\frac{3}{2}} \cdot \sqrt x </math><br><br>		4) <math> {\frac{d}{dx}} [x\sqrt x] = {\frac{d}{dx}} [x^{1} \cdot x^{{\frac{1}{2}}}] = {\frac{d}{dx}} [x^{{\frac{3}{2}}}] = {\frac{3}{2}} \cdot x^{{\frac{3}{2}} -1} = {\frac{3}{2x}}^{{\frac{3}{2}}} - {\frac{2}{2}} = {\frac{3}{2x}}^{{\frac{1}{2}}} = {\frac{3}{2}} \cdot \sqrt x </math><br><br>

	5) <math> {\frac{d}{dx}} [{\frac{3}{x}} + \sqrt 2 - 5^{x}] = 3 \cdot {\frac{d}{dx}} [{\frac{1}{x}}] = 3 \cdot {\frac{d}{dx}} [x^{-1}] = (-1)(3) \cdot x^{-1-1} = -3 \cdot x^{-2} = {\frac{3}{2x}} + 0 -ln(-5) \cdot -5^{x} = -{\frac{3}{x^{2}} - ln(-5) \cdot -5^{x} </math><br><br>		5) <math> {\frac{d}{dx}} [{\frac{3}{x}} + \sqrt 2 - 5^{x}] = 3 \cdot {\frac{d}{dx}} [{\frac{1}{x}}] = 3 \cdot {\frac{d}{dx}} [x^{-1}] = (-1)(3) \cdot x^{-1-1} = -3 \cdot x^{-2} = {\frac{3}{x^{2}}} + 0 - ln(-5) \cdot -5^{x} = -{\frac{3}{x^{2}}} - ln(-5) \cdot -5^{x} </math> <br><br>

			6) <math> {\frac{d}{dx}} [{\frac{4}{\sqrt[3]x^{2}}} - 3x^{9} + \sqrt[3]x \cdot {\frac{1}{x}}] = 4 \cdot x^{-{\frac{2}{3}}} = x^{-{\frac{2}{3}}-{\frac{3}{3}}} = x^{-{\frac{5}{3}}} </math><br><br>

Aroldog69360@students.laalliance.org at 21:05, 13 April 2023

2023-04-13T21:05:05Z

Aroldog69360@students.laalliance.org at 16:02, 13 April 2023

2023-04-13T16:02:42Z

← Older revision		Revision as of 16:02, 13 April 2023
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	2) <math>{\frac{d}{dx}} [3x] = 3 \cdot {\frac{d}{dx}} [x] = 3(1)(x)^{1-1} = 3 \cdot x^0 = 3 </math> <br><br>		2) <math>{\frac{d}{dx}} [3x] = 3 \cdot {\frac{d}{dx}} [x] = 3(1)(x)^{1-1} = 3 \cdot x^0 = 3 </math> <br><br>
	3) <math>{\frac{d}{dx}} [9x^2] = 18x </math> <br><br>		3) <math>{\frac{d}{dx}} [9x^2] = 18x </math> <br><br>
	4) <math>{\frac{d}{dx}} [x^3 + x^2 + 10] = 3x^2 +2x + 0 </math> <br>		4) <math>{\frac{d}{dx}} [x^3 + x^2 + 10] = 3x^2 +2x + 0 </math> <br><br>
	5) <math>{\frac{d}{dx}} [\sqrt(x)] = {\frac{d}{dx}} [x^{\frac{1}{2}}] = {\frac{1}{2x}}^{{\frac{1}{2}}-{\frac{2}{2}}} = {\frac{1}{2x}}^{\frac{-1}{2}} = {\frac{1}{2\sqrt(x)}}</math><br>		5) <math>{\frac{d}{dx}} [\sqrt(x)] = {\frac{d}{dx}} [x^{\frac{1}{2}}] = {\frac{1}{2x}}^{{\frac{1}{2}}-{\frac{2}{2}}} = {\frac{1}{2x}}^{\frac{-1}{2}} = {\frac{1}{2\sqrt(x)}}</math><br><br>
	A1. <math> {\sqrt[n](x^m)} = ({\sqrt[n]x})^m = x^{\frac{m}{n}} </math><br>		A1. <math> {\sqrt[n](x^m)} = ({\sqrt[n]x})^m = x^{\frac{m}{n}} </math><br><br>
	6) <math> {\frac{d}{dx}} [{\sqrt[3](x^2)}] = {\frac{d}{dx}} [x^{\frac{2}{3}}] = {\frac{2}{3}} \cdot x^{\frac{-1}{3}} = {\frac{2}{3} \cdot \sqrt[3](x)} </math><br>		6) <math> {\frac{d}{dx}} [{\sqrt[3](x^2)}] = {\frac{d}{dx}} [x^{\frac{2}{3}}] = {\frac{2}{3}} \cdot x^{\frac{-1}{3}} = {\frac{2}{3} \cdot \sqrt[3](x)} </math><br><br>
	7) <math> {\frac{d}{dx}} [(\sqrt[5]x)^7] = {\frac{d}{dx}} [{\frac{7}{x^5}}] = {\frac{7}{5}} \cdot x^{{\frac{7}{5}} - 1} = {\frac{7}{5}} \cdot x^{{\frac{2}{5}}} = {\frac{7}{5}} \cdot {\sqrt [5]x^2} </math><br>		7) <math> {\frac{d}{dx}} [(\sqrt[5]x)^7] = {\frac{d}{dx}} [{\frac{7}{x^5}}] = {\frac{7}{5}} \cdot x^{{\frac{7}{5}} - 1} = {\frac{7}{5}} \cdot x^{{\frac{2}{5}}} = {\frac{7}{5}} \cdot {\sqrt [5]x^2} </math><br><br>
	8) <math> {\frac{d}{dx}} [3x^{10}+e^{x}-5^{x}] = 30x^{9} + e^{x} - ln(5) \cdot 5^{x} </math><br>		8) <math> {\frac{d}{dx}} [3x^{10}+e^{x}-5^{x}] = 30x^{9} + e^{x} - ln(5) \cdot 5^{x} </math><br><br>
	9) <math> {\frac{d}{dx}} [{\frac{1}{x}}] = {\frac{d}{dx}} [x^{-1}] = -1 \cdot x^{-1-1} = -x^{-2} = - {\frac{1}{x^2}} </math><br>		9) <math> {\frac{d}{dx}} [{\frac{1}{x}}] = {\frac{d}{dx}} [x^{-1}] = -1 \cdot x^{-1-1} = -x^{-2} = - {\frac{1}{x^2}} </math><br><br><br>
			<math>\mathbf{\color{Purple}{Practice Problems}}</math><br>

			1) <math> {\frac{d}{dx}} [\pi + 3x -5x^{3}] = 0+3-15x^{2} </math><br><br>
			2) <math> {\frac{d}{dx}} [e^{x} + 3^{x} - 5 \cdot 2^{x}] = e^{x} + ln(3) \cdot 3^{x} -5 \cdot ln(2) \cdot 2^{x} </math><br><br>
			3) <math> {\frac{d}{dx}} [x^{5} - 6x^{3} +2x +5] = 5x^{4} -18x^{2} + 2 + 0 </math><br><br>
			4) <math> {\frac{d}{dx}} [x\sqrt x] = {\frac{d}{dx}} [x^{1} \cdot x^{{\frac{1}{2}}}] = {\frac{d}{dx}} [x^{{\frac{3}{2}}}] = {\frac{3}{2}} \cdot x^{{\frac{3}{2}}} -1} = {\frac{3}{2x}}^{{\frac{3}{2}}} - {\frac{2}{2}} = {\frac{3}{2x}}^{{\frac{1}{2}} = {\frac{3}{2}} \cdot \sqrt x </math><br><br>

Aroldog69360@students.laalliance.org at 18:09, 11 April 2023

2023-04-11T18:09:54Z

← Older revision		Revision as of 18:09, 11 April 2023
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	4) <math>{\frac{d}{dx}} [x^3 + x^2 + 10] = 3x^2 +2x + 0 </math> <br>		4) <math>{\frac{d}{dx}} [x^3 + x^2 + 10] = 3x^2 +2x + 0 </math> <br>
	5) <math>{\frac{d}{dx}} [\sqrt(x)] = {\frac{d}{dx}} [x^{\frac{1}{2}}] = {\frac{1}{2x}}^{{\frac{1}{2}}-{\frac{2}{2}}} = {\frac{1}{2x}}^{\frac{-1}{2}} = {\frac{1}{2\sqrt(x)}}</math><br>		5) <math>{\frac{d}{dx}} [\sqrt(x)] = {\frac{d}{dx}} [x^{\frac{1}{2}}] = {\frac{1}{2x}}^{{\frac{1}{2}}-{\frac{2}{2}}} = {\frac{1}{2x}}^{\frac{-1}{2}} = {\frac{1}{2\sqrt(x)}}</math><br>
	A1.		A1. <math> {\sqrt[n](x^m)} = ({\sqrt[n]x})^m = x^{\frac{m}{n}} </math><br>
			6) <math> {\frac{d}{dx}} [{\sqrt[3](x^2)}] = {\frac{d}{dx}} [x^{\frac{2}{3}}] = {\frac{2}{3}} \cdot x^{\frac{-1}{3}} = {\frac{2}{3} \cdot \sqrt[3](x)} </math><br>
	<math> \sqrt(x) </math>		7) <math> {\frac{d}{dx}} [(\sqrt[5]x)^7] = {\frac{d}{dx}} [{\frac{7}{x^5}}] = {\frac{7}{5}} \cdot x^{{\frac{7}{5}} - 1} = {\frac{7}{5}} \cdot x^{{\frac{2}{5}}} = {\frac{7}{5}} \cdot {\sqrt [5]x^2} </math><br>
			8) <math> {\frac{d}{dx}} [3x^{10}+e^{x}-5^{x}] = 30x^{9} + e^{x} - ln(5) \cdot 5^{x} </math><br>
	\sqrt{x} <br><br><br><br><br><br><br><br><br>		9) <math> {\frac{d}{dx}} [{\frac{1}{x}}] = {\frac{d}{dx}} [x^{-1}] = -1 \cdot x^{-1-1} = -x^{-2} = - {\frac{1}{x^2}} </math><br>

Aroldog69360@students.laalliance.org at 17:38, 11 April 2023

2023-04-11T17:38:47Z

← Older revision		Revision as of 17:38, 11 April 2023
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	3) <math>{\frac{d}{dx}} [9x^2] = 18x </math> <br><br>		3) <math>{\frac{d}{dx}} [9x^2] = 18x </math> <br><br>
	4) <math>{\frac{d}{dx}} [x^3 + x^2 + 10] = 3x^2 +2x + 0 </math> <br>		4) <math>{\frac{d}{dx}} [x^3 + x^2 + 10] = 3x^2 +2x + 0 </math> <br>
	5) <math>{\frac{d}{dx}} [\sqrt(x)] = {\frac{d}{dx}} [x^{{\frac{1}{2}}] = {\frac{1}{2x}^{{\frac{1}{2}-1}} </math> <br>		5) <math>{\frac{d}{dx}} [\sqrt(x)] = {\frac{d}{dx}} [x^{\frac{1}{2}}] = {\frac{1}{2x}}^{{\frac{1}{2}}-{\frac{2}{2}}} = {\frac{1}{2x}}^{\frac{-1}{2}} = {\frac{1}{2\sqrt(x)}}</math><br>
			A1.

			<math> \sqrt(x) </math>

			\sqrt{x} <br><br><br><br><br><br><br><br><br>
	\sqrt{x}



	<br><br><br><br><br><br><br><br><br>
	<math>{\frac{d}{dx}} [c] = 0 </math> <br> <math>{\frac{d}{dx}} [c\cdot f(x)] = c\cdot{\frac{d}{dx}} [f(x)] </math> <br> <math>{\frac{d}{dx}} [f(x)\pm g(x)] = {\frac{d}{dx}} [f(x)] \pm {\frac{d}{dx}} [g(x)] </math> <br> <math>{\frac{d}{dx}} [a^x] = \ln(a)a^x </math><br> <math>{\frac{d}{dx}} [e^x] = e^x </math><br> <math>\color{Blue}Power\,Rule </math><br> <math>{\frac{d}{dx}} [x^n] = n \cdot x^n-1 </math> <br> <math>\color{Red}Product\,Rule </math><br> <math>{\frac{d}{dx}} [f\cdot{g}]= {\frac{d}{dx}}[f]\cdot{g}+{\frac{d}{dx}}[g]\cdot{f}</math><br> <math>\color{Green}Quotient\,Rule </math><br> <math>{\frac{d}{dx}}[\frac{f}{g}]=\frac{{\frac{d}{dx}}[f]\cdot{g}-{\frac{d}{dx}}[g]\cdot{f}}{g^2}</math><br> <math>\mathbf{\color{Purple}{Examples}}</math><br> <math>\mathbf{Ex.1}</math><br> <math>if\,f(x)=x\cdot{e^x}</math><br> <math>f^\prime(x)=1\cdot{e^x}+x\cdot{e^x}</math><br> <math>\mathbf{Ex.2}</math><br> <math>if\,f(t)=\sqrt{t}(a+bt)</math><br> <math>f^\prime(t)=\frac{1}{2\sqrt{t}}(a+bt)+t\sqrt{t}(b)</math><br> <math>\mathbf{Ex.3}</math><br> <math>if\,f(x)=\sqrt{x}\cdot{g(x)}</math><br> <math>g(4)=2</math><br> <math>g^\prime(4)=3</math><br> <math>f^\prime(x)=\frac{1}{2\sqrt{x}}\cdot{g(x)}+\sqrt{x}\cdot{g^\prime(x)}</math><br> <math>\mathbf{Ex.4}</math><br> <math>y=\frac{\color{Blue}{x^2+x-2}}{\color{Red}{x^3+6}}</math><br> <math>{\frac{d}{dx}}=y^\prime=\frac{(2x+1)(x^3-6)-\color{Blue}{(x^2+x-2)}(3x^2)}{\color{Red}{(x^3+6)^2}}</math><br> <math>=\frac{(2x^4+x^4+x^3+12x+6-[3x^4+3x^2-6x^2]}{(x^3+6)^2}</math><br> <math>=\frac{-x^4-2x^3+6x^2+12x+6}{(x^3+6)^2}</math><br> <math>\mathbf{Ex.5}</math><br> <math>y=\frac{e^x}{1+x^2}\,(1,\frac{e}{2})\,</math><br> <math>{\frac{d}{dx}}=\frac{e^x\cdot(1+x^2)-e^x(2x)}{(1+x^2)^2}</math><br> <math>{\frac{d}{dx}}\|_{x=1}\frac{e(1+1)-e^\prime(2)}{(1+1)^2}=\frac{2e-2e}{2^2}=\frac{0}{4}=0</math>

Aroldog69360@students.laalliance.org at 17:08, 11 April 2023

2023-04-11T17:08:46Z

← Older revision		Revision as of 17:08, 11 April 2023
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	3) <math>{\frac{d}{dx}} [9x^2] = 18x </math> <br><br>		3) <math>{\frac{d}{dx}} [9x^2] = 18x </math> <br><br>
	4) <math>{\frac{d}{dx}} [x^3 + x^2 + 10] = 3x^2 +2x + 0 </math> <br>		4) <math>{\frac{d}{dx}} [x^3 + x^2 + 10] = 3x^2 +2x + 0 </math> <br>
	5) <math>{\frac{d}{dx}} [sqrt(x)] = {\frac{d}{dx}} [x^{{\frac{1}{2}}] = {\frac{1}{2x}^{{\frac{1}{2}-1}} </math> <br>		5) <math>{\frac{d}{dx}} [\sqrt(x)] = {\frac{d}{dx}} [x^{{\frac{1}{2}}] = {\frac{1}{2x}^{{\frac{1}{2}-1}} </math> <br>

Aroldog69360@students.laalliance.org at 17:08, 11 April 2023

2023-04-11T17:08:27Z

← Older revision		Revision as of 17:08, 11 April 2023
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	3) <math>{\frac{d}{dx}} [9x^2] = 18x </math> <br><br>		3) <math>{\frac{d}{dx}} [9x^2] = 18x </math> <br><br>
	4) <math>{\frac{d}{dx}} [x^3 + x^2 + 10] = 3x^2 +2x + 0 </math> <br>		4) <math>{\frac{d}{dx}} [x^3 + x^2 + 10] = 3x^2 +2x + 0 </math> <br>
	5) <math>{\frac{d}{dx}} [√x] = {\frac{d}{dx}} [x^{{\frac{1}{2}}] = {\frac{1}{2x}^{{\frac{1}{2}-1}} </math> <br>		5) <math>{\frac{d}{dx}} [sqrt(x)] = {\frac{d}{dx}} [x^{{\frac{1}{2}}] = {\frac{1}{2x}^{{\frac{1}{2}-1}} </math> <br>

Aroldog69360@students.laalliance.org: Created page with " $\mathbf{3.1}$
1. The derivative of a constant is 0 = ${\frac{d}{dx}} [c] = 0$
2. ${\frac{d}{dx}} [x^n] = nx^{n-1}$

3. ${\frac{d}{dx}} [c\cdot f(x)] = c \cdot {\frac{d} {dx}} [f(x)]$

4. ${\frac{d}{dx}} [f(x) + g(x)] = {\frac{d}{dx}} [f(x)] + {\frac{d}{dx}} [g(x)]$

5. ${\frac{d}{dx}} [a^x] = ln (a) \cdot a^x$

6. ${\frac{d}{dx}} [e^x] = e^x$
2023-03-30T22:20:47Z

Created page with "<math>\mathbf{3.1}</math> 1. The derivative of a constant is 0 = <math>{\frac{d}{dx}} [c] = 0 </math> 2. <math>{\frac{d}{dx}} [x^n] = nx^{n-1} </math> 3. <math>{\frac{d}{dx}} [c\cdot f(x)] = c \cdot {\frac{d} {dx}} [f(x)] </math> 4. <math>{\frac{d}{dx}} [f(x) + g(x)] = {\frac{d}{dx}} [f(x)] + {\frac{d}{dx}} [g(x)] </math> 5. <math>{\frac{d}{dx}} [a^x] = ln (a) \cdot a^x </math> 6. <math>{\frac{d}{dx}} [e^x] = e^x </math> <br..."

New page
<math>\mathbf{3.1}</math> 
1. The derivative of a constant is 0 = <math>{\frac{d}{dx}} [c] = 0 </math> 
2. <math>{\frac{d}{dx}} [x^n] = nx^{n-1} </math> 
3. <math>{\frac{d}{dx}} [c\cdot f(x)] = c \cdot {\frac{d} {dx}} [f(x)] </math> 
4. <math>{\frac{d}{dx}} [f(x) + g(x)] = {\frac{d}{dx}} [f(x)] + {\frac{d}{dx}} [g(x)] </math> 
5. <math>{\frac{d}{dx}} [a^x] = ln (a) \cdot a^x </math> 
6. <math>{\frac{d}{dx}} [e^x] = e^x </math> 
<math>\mathbf{\color{Blue}{Examples}}</math> 

1) <math>{\frac{d}{dx}} [5 + \pi] = 0+0=0 </math> 
2) <math>{\frac{d}{dx}} [3x] = 3 \cdot {\frac{d}{dx}} [x] = 3(1)(x)^{1-1} = 3 \cdot x^0 = 3 </math> 
3) <math>{\frac{d}{dx}} [9x^2] = 18x </math> 
4) <math>{\frac{d}{dx}} [x^3 + x^2 + 10] = 3x^2 +2x + 0 </math> 
5) <math>{\frac{d}{dx}} [√x] = {\frac{d}{dx}} [x^{{\frac{1}{2}}] = {\frac{1}{2x}^{{\frac{1}{2}-1}} </math> 

\sqrt{x}

 
<math>{\frac{d}{dx}} [c] = 0 </math> <math>{\frac{d}{dx}} [c\cdot f(x)] = c\cdot{\frac{d}{dx}} [f(x)] </math> <math>{\frac{d}{dx}} [f(x)\pm g(x)] = {\frac{d}{dx}} [f(x)] \pm {\frac{d}{dx}} [g(x)] </math> <math>{\frac{d}{dx}} [a^x] = \ln(a)a^x </math> <math>{\frac{d}{dx}} [e^x] = e^x </math> <math>\color{Blue}Power\,Rule </math> <math>{\frac{d}{dx}} [x^n] = n \cdot x^n-1 </math> <math>\color{Red}Product\,Rule </math> <math>{\frac{d}{dx}} [f\cdot{g}]= {\frac{d}{dx}}[f]\cdot{g}+{\frac{d}{dx}}[g]\cdot{f}</math> <math>\color{Green}Quotient\,Rule </math> <math>{\frac{d}{dx}}[\frac{f}{g}]=\frac{{\frac{d}{dx}}[f]\cdot{g}-{\frac{d}{dx}}[g]\cdot{f}}{g^2}</math> <math>\mathbf{\color{Purple}{Examples}}</math> <math>\mathbf{Ex.1}</math> <math>if\,f(x)=x\cdot{e^x}</math> <math>f^\prime(x)=1\cdot{e^x}+x\cdot{e^x}</math> <math>\mathbf{Ex.2}</math> <math>if\,f(t)=\sqrt{t}(a+bt)</math> <math>f^\prime(t)=\frac{1}{2\sqrt{t}}(a+bt)+t\sqrt{t}(b)</math> <math>\mathbf{Ex.3}</math> <math>if\,f(x)=\sqrt{x}\cdot{g(x)}</math> <math>g(4)=2</math> <math>g^\prime(4)=3</math> <math>f^\prime(x)=\frac{1}{2\sqrt{x}}\cdot{g(x)}+\sqrt{x}\cdot{g^\prime(x)}</math> <math>\mathbf{Ex.4}</math> <math>y=\frac{\color{Blue}{x^2+x-2}}{\color{Red}{x^3+6}}</math> <math>{\frac{d}{dx}}=y^\prime=\frac{(2x+1)(x^3-6)-\color{Blue}{(x^2+x-2)}(3x^2)}{\color{Red}{(x^3+6)^2}}</math> <math>=\frac{(2x^4+x^4+x^3+12x+6-[3x^4+3x^2-6x^2]}{(x^3+6)^2}</math> <math>=\frac{-x^4-2x^3+6x^2+12x+6}{(x^3+6)^2}</math> <math>\mathbf{Ex.5}</math> <math>y=\frac{e^x}{1+x^2}\,(1,\frac{e}{2})\,</math> <math>{\frac{d}{dx}}=\frac{e^x\cdot(1+x^2)-e^x(2x)}{(1+x^2)^2}</math> <math>{\frac{d}{dx}}|_{x=1}\frac{e(1+1)-e^\prime(2)}{(1+1)^2}=\frac{2e-2e}{2^2}=\frac{0}{4}=0</math>

← Older revision		Revision as of 21:05, 13 April 2023
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	1) <math>{\frac{d}{dx}} [5 + \pi] = 0+0=0 </math> <br><br>		1) <math>{\frac{d}{dx}} [5 + \pi] = 0+0=0 </math> <br><br>

	2) <math>{\frac{d}{dx}} [3x] = 3 \cdot {\frac{d}{dx}} [x] = 3(1)(x)^{1-1} = 3 \cdot x^0 = 3 </math> <br><br>		2) <math>{\frac{d}{dx}} [3x] = 3 \cdot {\frac{d}{dx}} [x] = 3(1)(x)^{1-1} = 3 \cdot x^0 = 3 </math> <br><br>

	3) <math>{\frac{d}{dx}} [9x^2] = 18x </math> <br><br>		3) <math>{\frac{d}{dx}} [9x^2] = 18x </math> <br><br>

	4) <math>{\frac{d}{dx}} [x^3 + x^2 + 10] = 3x^2 +2x + 0 </math> <br><br>		4) <math>{\frac{d}{dx}} [x^3 + x^2 + 10] = 3x^2 +2x + 0 </math> <br><br>

	5) <math>{\frac{d}{dx}} [\sqrt(x)] = {\frac{d}{dx}} [x^{\frac{1}{2}}] = {\frac{1}{2x}}^{{\frac{1}{2}}-{\frac{2}{2}}} = {\frac{1}{2x}}^{\frac{-1}{2}} = {\frac{1}{2\sqrt(x)}}</math><br><br>		5) <math>{\frac{d}{dx}} [\sqrt(x)] = {\frac{d}{dx}} [x^{\frac{1}{2}}] = {\frac{1}{2x}}^{{\frac{1}{2}}-{\frac{2}{2}}} = {\frac{1}{2x}}^{\frac{-1}{2}} = {\frac{1}{2\sqrt(x)}}</math><br><br>

	A1. <math> {\sqrt[n](x^m)} = ({\sqrt[n]x})^m = x^{\frac{m}{n}} </math><br><br>		A1. <math> {\sqrt[n](x^m)} = ({\sqrt[n]x})^m = x^{\frac{m}{n}} </math><br><br>

	6) <math> {\frac{d}{dx}} [{\sqrt[3](x^2)}] = {\frac{d}{dx}} [x^{\frac{2}{3}}] = {\frac{2}{3}} \cdot x^{\frac{-1}{3}} = {\frac{2}{3} \cdot \sqrt[3](x)} </math><br><br>		6) <math> {\frac{d}{dx}} [{\sqrt[3](x^2)}] = {\frac{d}{dx}} [x^{\frac{2}{3}}] = {\frac{2}{3}} \cdot x^{\frac{-1}{3}} = {\frac{2}{3} \cdot \sqrt[3](x)} </math><br><br>

	7) <math> {\frac{d}{dx}} [(\sqrt[5]x)^7] = {\frac{d}{dx}} [{\frac{7}{x^5}}] = {\frac{7}{5}} \cdot x^{{\frac{7}{5}} - 1} = {\frac{7}{5}} \cdot x^{{\frac{2}{5}}} = {\frac{7}{5}} \cdot {\sqrt [5]x^2} </math><br><br>		7) <math> {\frac{d}{dx}} [(\sqrt[5]x)^7] = {\frac{d}{dx}} [{\frac{7}{x^5}}] = {\frac{7}{5}} \cdot x^{{\frac{7}{5}} - 1} = {\frac{7}{5}} \cdot x^{{\frac{2}{5}}} = {\frac{7}{5}} \cdot {\sqrt [5]x^2} </math><br><br>

	8) <math> {\frac{d}{dx}} [3x^{10}+e^{x}-5^{x}] = 30x^{9} + e^{x} - ln(5) \cdot 5^{x} </math><br><br>		8) <math> {\frac{d}{dx}} [3x^{10}+e^{x}-5^{x}] = 30x^{9} + e^{x} - ln(5) \cdot 5^{x} </math><br><br>

	9) <math> {\frac{d}{dx}} [{\frac{1}{x}}] = {\frac{d}{dx}} [x^{-1}] = -1 \cdot x^{-1-1} = -x^{-2} = - {\frac{1}{x^2}} </math><br><br><br>		9) <math> {\frac{d}{dx}} [{\frac{1}{x}}] = {\frac{d}{dx}} [x^{-1}] = -1 \cdot x^{-1-1} = -x^{-2} = - {\frac{1}{x^2}} </math><br><br><br>

	<math>\mathbf{\color{Purple}{Practice Problems}}</math><br>		<math>\mathbf{\color{Purple}{Practice Problems}}</math><br>

	1) <math> {\frac{d}{dx}} [\pi + 3x -5x^{3}] = 0+3-15x^{2} </math><br><br>		1) <math> {\frac{d}{dx}} [\pi + 3x -5x^{3}] = 0+3-15x^{2} </math><br><br>

	2) <math> {\frac{d}{dx}} [e^{x} + 3^{x} - 5 \cdot 2^{x}] = e^{x} + ln(3) \cdot 3^{x} -5 \cdot ln(2) \cdot 2^{x} </math><br><br>		2) <math> {\frac{d}{dx}} [e^{x} + 3^{x} - 5 \cdot 2^{x}] = e^{x} + ln(3) \cdot 3^{x} -5 \cdot ln(2) \cdot 2^{x} </math><br><br>

	3) <math> {\frac{d}{dx}} [x^{5} - 6x^{3} +2x +5] = 5x^{4} -18x^{2} + 2 + 0 </math><br><br>		3) <math> {\frac{d}{dx}} [x^{5} - 6x^{3} +2x +5] = 5x^{4} -18x^{2} + 2 + 0 </math><br><br>
	4) <math> {\frac{d}{dx}} [x\sqrt x] = {\frac{d}{dx}} [x^{1} \cdot x^{{\frac{1}{2}}}] = {\frac{d}{dx}} [x^{{\frac{3}{2}}}] = {\frac{3}{2}} \cdot x^{{\frac{3}{2}}} -1} = {\frac{3}{2x}}^{{\frac{3}{2}}} - {\frac{2}{2}} = {\frac{3}{2x}}^{{\frac{1}{2}} = {\frac{3}{2}} \cdot \sqrt x </math><br><br>
			4) <math> {\frac{d}{dx}} [x\sqrt x] = {\frac{d}{dx}} [x^{1} \cdot x^{{\frac{1}{2}}}] = {\frac{d}{dx}} [x^{{\frac{3}{2}}}] = {\frac{3}{2}} \cdot x^{{\frac{3}{2}} -1} = {\frac{3}{2x}}^{{\frac{3}{2}}} - {\frac{2}{2}} = {\frac{3}{2x}}^{{\frac{1}{2}}} = {\frac{3}{2}} \cdot \sqrt x </math><br><br>

			5) <math> {\frac{d}{dx}} [{\frac{3}{x}} + \sqrt 2 - 5^{x}] = 3 \cdot {\frac{d}{dx}} [{\frac{1}{x}}] = 3 \cdot {\frac{d}{dx}} [x^{-1}] = (-1)(3) \cdot x^{-1-1} = -3 \cdot x^{-2} = {\frac{3}{2x}} + 0 -ln(-5) \cdot -5^{x} = -{\frac{3}{x^{2}} - ln(-5) \cdot -5^{x} </math><br><br>