2024/G9/12 - Revision history

Freddys70014@students.laalliance.org at 16:17, 30 March 2023

2023-03-30T16:17:47Z

← Older revision		Revision as of 16:17, 30 March 2023
Line 1:		Line 1:
	~~qQ`~~<math>\mathbf{Chapter 3 Section 2}</math><br>		<math>\mathbf{Chapter 3 Section 2}</math><br>
	<math>{\frac{d}{dx}} [c] = 0 </math> <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}} [c\cdot f(x)] = c\cdot{\frac{d}{dx}} [f(x)] </math> <br>
Line 11:		Line 11:
	<math>\color{Green}Quotient\,Rule </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>{\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>\mathbf{Ex.1}</math><br>
	<math>if\,f(x)=x\cdot{e^x}</math><br>		<math>if\,f(x)=x\cdot{e^x}</math><br>
Line 30:		Line 31:
	<math>y=\frac{e^x}{1+x^2}\,(1,\frac{e}{2})\,</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}}=\frac{e^x\cdot(1+x^2)-e^x(2x)}{(1+x^2)^2}</math><br>
	<math>{\frac{d}{dx}}		<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>

Freddys70014@students.laalliance.org at 15:52, 30 March 2023

2023-03-30T15:52:05Z

← Older revision		Revision as of 15:52, 30 March 2023
Line 1:		Line 1:
	<math>\mathbf{Chapter 3 Section 2}</math><br>		qQ`<math>\mathbf{Chapter 3 Section 2}</math><br>
	<math>{\frac{d}{dx}} [c] = 0 </math> <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}} [c\cdot f(x)] = c\cdot{\frac{d}{dx}} [f(x)] </math> <br>
Line 27:		Line 27:
	<math>=\frac{(2x^4+x^4+x^3+12x+6-[3x^4+3x^2-6x^2]}{(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>=\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}}

Freddys70014@students.laalliance.org at 18:13, 29 March 2023

2023-03-29T18:13:13Z

← Older revision		Revision as of 18:13, 29 March 2023
Line 23:		Line 23:
	<math>f^\prime(x)=\frac{1}{2\sqrt{x}}\cdot{g(x)}+\sqrt{x}\cdot{g^\prime(x)}</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>\mathbf{Ex.4}</math><br>
	<math>y=\frac{x^2+x-2}{x^3+6}</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)-(x^2+x-2)(3x^2)}{(x^3+6)^2}</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{(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>=\frac{-x^4-2x^3+6x^2+12x+6}{(x^3+6)^2}</math><br>

Freddys70014@students.laalliance.org at 18:08, 29 March 2023

2023-03-29T18:08:21Z

← Older revision		Revision as of 18:08, 29 March 2023
Line 5:		Line 5:
	<math>{\frac{d}{dx}} [a^x] = \ln(a)a^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>{\frac{d}{dx}} [e^x] = e^x </math><br>
	<math>\color{Blue}Power Rule </math><br>		<math>\color{Blue}Power\,Rule </math><br>
	<math>{\frac{d}{dx}} [x^n] = n \cdot x^n-1 </math> <br>		<math>{\frac{d}{dx}} [x^n] = n \cdot x^n-1 </math> <br>
	<math>\color{~~Blue~~}Product Rule </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>{\frac{d}{dx}} [f\cdot{g}]= {\frac{d}{dx}}[f]\cdot{g}+{\frac{d}{dx}}[g]\cdot{f}</math><br>
	<math>\color{~~Blue~~}Quotient Rule </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>{\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{Ex.1}</math><br>		<math>\mathbf{Ex.1}</math><br>
	<math>f(x)=x\cdot{e^x}</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>f^\prime(x)=1\cdot{e^x}+x\cdot{e^x}</math><br>
	<math>\mathbf{Ex.2}</math><br>		<math>\mathbf{Ex.2}</math><br>
	<math>f(t)=\sqrt{t}(a+bt)</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>f^\prime(t)=\frac{1}{2\sqrt{t}}(a+bt)+t\sqrt{t}(b)</math><br>
	<math>\mathbf{Ex.3}</math><br>		<math>\mathbf{Ex.3}</math><br>
	<math>if\,f(x)=\sqrt{x}\cdot{g(x)}</math><br>		<math>if\,f(x)=\sqrt{x}\cdot{g(x)}</math><br>
	<math>g(4)=2</math><br>		<math>g(4)=2</math><br>
	<math>g^\prime(4)=3</math>		<math>g^\prime(4)=3</math><br>
	<math>f^\prime(x)=1\~~sqrt~~{1}{2\sqrt{x}}\cdot{g(x)}+\sqrt{x}\cdot{g^\prime(x)}</math>		<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{x^2+x-2}{x^3+6}</math><br>
			<math>{\frac{d}{dx}}=y^\prime=\frac{(2x+1)(x^3-6)-(x^2+x-2)(3x^2)}{(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>

Freddys70014@students.laalliance.org at 17:52, 29 March 2023

2023-03-29T17:52:46Z

← Older revision		Revision as of 17:52, 29 March 2023
Line 12:		Line 12:
	<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>{\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{Ex.1}</math><br>		<math>\mathbf{Ex.1}</math><br>
	a)<math>f(x)=x\cdot{e^x}</math><br>		<math>f(x)=x\cdot{e^x}</math><br>
	<math>f^\prime(x)=1\cdot{e^x}+x\cdot{e^x}		<math>f^\prime(x)=1\cdot{e^x}+x\cdot{e^x}</math><br>
	</math>		<math>\mathbf{Ex.2}</math><br>
			<math>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>
			<math>f^\prime(x)=1\sqrt{1}{2\sqrt{x}}\cdot{g(x)}+\sqrt{x}\cdot{g^\prime(x)}</math>

Freddys70014@students.laalliance.org at 23:11, 28 March 2023

2023-03-28T23:11:11Z

← Older revision		Revision as of 23:11, 28 March 2023
Line 11:		Line 11:
	<math>\color{Blue}Quotient Rule </math><br>		<math>\color{Blue}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>{\frac{d}{dx}}[\frac{f}{g}]=\frac{{\frac{d}{dx}}[f]\cdot{g}-{\frac{d}{dx}}[g]\cdot{f}}{g^2}</math><br>
	Ex.1<br>		<math>\mathbf{Ex.1}</math><br>
	a)<math>f(x)=x\cdot{e^x}</math><br>		a)<math>f(x)=x\cdot{e^x}</math><br>
	<math>f^\prime(x)=1\cdot{e^x}+x\cdot{e^x}		<math>f^\prime(x)=1\cdot{e^x}+x\cdot{e^x}
	</math>		</math>

Freddys70014@students.laalliance.org at 23:09, 28 March 2023

2023-03-28T23:09:29Z

← Older revision		Revision as of 23:09, 28 March 2023
Line 1:		Line 1:
	Chapter 3 Section 2 ~~Notes~~<br>		<math>\mathbf{Chapter 3 Section 2}</math><br>
	<math>{\frac{d}{dx}} [c] = 0 </math> <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}} [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}} [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>\color{Blue}Power Rule </math><br>
	<math>{\frac{d}{dx}} [x^n] = n \cdot x^n-1 </math> <br>		<math>{\frac{d}{dx}} [x^n] = n \cdot x^n-1 </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}Product Rule </math><br>		<math>\color{Blue}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>{\frac{d}{dx}} [f\cdot{g}]= {\frac{d}{dx}}[f]\cdot{g}+{\frac{d}{dx}}[g]\cdot{f}</math><br>
	<math>\color{Blue}Quotient Rule </math><br>		<math>\color{Blue}Quotient Rule </math><br>
	<math>{\frac{d}{dx}}[/frac{f}{g}]=/frac{{\frac{d}{dx}}[f]\cdot{g}-{\frac{d}{dx}}[g]\cdot{f}}</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>
			Ex.1<br>
			a)<math>f(x)=x\cdot{e^x}</math><br>
			<math>f^\prime(x)=1\cdot{e^x}+x\cdot{e^x}
			</math>

Freddys70014@students.laalliance.org at 22:55, 28 March 2023

2023-03-28T22:55:47Z

← Older revision		Revision as of 22:55, 28 March 2023
Line 1:		Line 1:
			Chapter 3 Section 2 Notes<br>
	<math>{\frac{d}{dx}} [c] = 0 </math> <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}} [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}} [f(x)\pm g(x)] = {\frac{d}{dx}} [f(x)] \pm {\frac{d}{dx}} [g(x)] </math> <br>
	<math> {\frac{d}{dx}} [x^n] = n \cdot x^n-1 </math> <br>		<math>\color{Blue}Power Rule </math><br>
			<math>{\frac{d}{dx}} [x^n] = n \cdot x^n-1 </math> <br>
	<math>{\frac{d}{dx}} [a^x] = \ln(a)a^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>{\frac{d}{dx}} [e^x] = e^x </math><br>
			<math>\color{Blue}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{Blue}Quotient Rule </math><br>
			<math>{\frac{d}{dx}}[/frac{f}{g}]=/frac{{\frac{d}{dx}}[f]\cdot{g}-{\frac{d}{dx}}[g]\cdot{f}}</math><br>

Freddys70014@students.laalliance.org: Created page with " ${\frac{d}{dx}} [c] = 0$
${\frac{d}{dx}} [c\cdot f(x)] = c\cdot{\frac{d}{dx}} [f(x)]$
${\frac{d}{dx}} [f(x)\pm g(x)] = {\frac{d}{dx}} [f(x)] \pm {\frac{d}{dx}} [g(x)]$
${\frac{d}{dx}} [x^n] = n \cdot x^n-1$
${\frac{d}{dx}} [a^x] = \ln(a)a^x$
${\frac{d}{dx}} [e^x] = e^x$
"

2023-03-28T20:13:38Z

Created page with "<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}} [x^n] = n \cdot x^n-1 </math> <math>{\frac{d}{dx}} [a^x] = \ln(a)a^x </math> <math> {\frac{d}{dx}} [e^x] = e^x </math> "

New page