2024/G2/3: Difference between revisions

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<math>\mathbf{Chapter 3 Section 2}</math><br>
[[2024/G9/12|Section 2]]<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>

Revision as of 21:40, 30 March 2023

Section 2

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