Freddys70014@students.laalliance.org: Created page with " $\mathbf{Chapter 3 Section 2}$
${\frac{d}{dx}} [c] = 0$
${\frac{d}{dx}} [c\cdot f(x)] = c\cdot{\frac{d}{dx}} [f(x)]$
${\frac{d}{dx}} [f(x)\pm g(x)] = {\frac{d}{dx}} [f(x)] \pm {\frac{d}{dx}} [g(x)]$
${\frac{d}{dx}} [a^x] = \ln(a)a^x$
${\frac{d}{dx}} [e^x] = e^x$
$\color{Blue}Power\,Rule$
${\frac{d}{dx}} [x^n] = n \cdot x^n-1$
$\color{Re..."$

2023-03-29T18:09:55Z

Created page with "<math>\mathbf{Chapter 3 Section 2}</math> <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{Re..."

New page

<math>\mathbf{Chapter 3 Section 2}</math> 
<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{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{x^2+x-2}{x^3+6}</math> 
<math>{\frac{d}{dx}}=y^\prime=\frac{(2x+1)(x^3-6)-(x^2+x-2)(3x^2)}{(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>

2024/G10/13 - Revision history