2024/G9/12: Difference between revisions

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<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>

Revision as of 18:08, 29 March 2023