2024/G2/12: Difference between revisions

Latest revision as of 16:45, 14 April 2023

$\mathbf {3.1}$
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. ${\frac {d}{dx}}[e^{x}]=e^{x}$

$\mathbf {\color {Blue}{Examples}}$

1) ${\frac {d}{dx}}[5+\pi ]=0+0=0$

2) ${\frac {d}{dx}}[3x]=3\cdot {\frac {d}{dx}}[x]=3(1)(x)^{1-1}=3\cdot x^{0}=3$

3) ${\frac {d}{dx}}[9x^{2}]=18x$

4) ${\frac {d}{dx}}[x^{3}+x^{2}+10]=3x^{2}+2x+0$

5) ${\frac {d}{dx}}[{\sqrt {(}}x)]={\frac {d}{dx}}[x^{\frac {1}{2}}]={\frac {1}{2x}}^{{\frac {1}{2}}-{\frac {2}{2}}}={\frac {1}{2x}}^{\frac {-1}{2}}={\frac {1}{2{\sqrt {(}}x)}}$

A1. ${{\sqrt[{n}]{(}}x^{m})}=({\sqrt[{n}]{x}})^{m}=x^{\frac {m}{n}}$

6) ${\frac {d}{dx}}[{{\sqrt[{3}]{(}}x^{2})}]={\frac {d}{dx}}[x^{\frac {2}{3}}]={\frac {2}{3}}\cdot x^{\frac {-1}{3}}={{\frac {2}{3}}\cdot {\sqrt[{3}]{(}}x)}$

7) ${\frac {d}{dx}}[({\sqrt[{5}]{x}})^{7}]={\frac {d}{dx}}[{\frac {7}{x^{5}}}]={\frac {7}{5}}\cdot x^{{\frac {7}{5}}-1}={\frac {7}{5}}\cdot x^{\frac {2}{5}}={\frac {7}{5}}\cdot {{\sqrt[{5}]{x}}^{2}}$

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

9) ${\frac {d}{dx}}[{\frac {1}{x}}]={\frac {d}{dx}}[x^{-1}]=-1\cdot x^{-1-1}=-x^{-2}=-{\frac {1}{x^{2}}}$

$\mathbf {\color {Purple}{PracticeProblems}}$

1) ${\frac {d}{dx}}[\pi +3x-5x^{3}]=0+3-15x^{2}$

2) ${\frac {d}{dx}}[e^{x}+3^{x}-5\cdot 2^{x}]=e^{x}+ln(3)\cdot 3^{x}-5\cdot ln(2)\cdot 2^{x}$

3) ${\frac {d}{dx}}[x^{5}-6x^{3}+2x+5]=5x^{4}-18x^{2}+2+0$

4) ${\frac {d}{dx}}[x{\sqrt {x}}]={\frac {d}{dx}}[x^{1}\cdot x^{\frac {1}{2}}]={\frac {d}{dx}}[x^{\frac {3}{2}}]={\frac {3}{2}}\cdot x^{{\frac {3}{2}}-1}={\frac {3}{2x}}^{\frac {3}{2}}-{\frac {2}{2}}={\frac {3}{2x}}^{\frac {1}{2}}={\frac {3}{2}}\cdot {\sqrt {x}}$

5) ${\frac {d}{dx}}[{\frac {3}{x}}+{\sqrt {2}}-5^{x}]=3\cdot {\frac {d}{dx}}[{\frac {1}{x}}]=3\cdot {\frac {d}{dx}}[x^{-1}]=(-1)(3)\cdot x^{-1-1}=-3\cdot x^{-2}={\frac {3}{x^{2}}}+0-ln(-5)\cdot -5^{x}=-{\frac {3}{x^{2}}}-ln(-5)\cdot -5^{x}$

6) ${\frac {d}{dx}}[{\frac {4}{{\sqrt[{3}]{x}}^{2}}}-3x^{9}+{\sqrt[{3}]{x}}\cdot {\frac {1}{x}}]=4\cdot x^{-{\frac {2}{3}}}=x^{-{\frac {2}{3}}-{\frac {3}{3}}}=x^{-{\frac {5}{3}}}$

@@ Line 9: / Line 9: @@
 ) <math>{\frac{d}{dx}} [5 + \pi] = 0+0=0 </math> <br><br>
 ) <math>{\frac{d}{dx}} [3x] = 3 \cdot {\frac{d}{dx}} [x] = 3(1)(x)^{1-1} = 3 \cdot x^0 = 3 </math> <br><br>
 ) <math>{\frac{d}{dx}} [9x^2] = 18x </math> <br><br>
-) <math>{\frac{d}{dx}} [x^3 + x^2 + 10] = 3x^2 +2x + 0 </math> <br>
-) <math>{\frac{d}{dx}} [\sqrt(x)] = {\frac{d}{dx}} [x^{{\frac{1}{2}}] = {\frac{1}{2x}^{{\frac{1}{2}-1}} </math> <br>
+) <math>{\frac{d}{dx}} [x^3 + x^2 + 10] = 3x^2 +2x + 0 </math> <br><br>
+) <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>
+) <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>
+) <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>
+) <math> {\frac{d}{dx}} [3x^{10}+e^{x}-5^{x}] = 30x^{9} + e^{x} - ln(5) \cdot 5^{x} </math><br><br>
+) <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}{PracticeProblems}}</math><br>
+) <math> {\frac{d}{dx}} [\pi + 3x -5x^{3}] = 0+3-15x^{2} </math><br><br>
+) <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>
-\sqrt{x}
+) <math> {\frac{d}{dx}} [x^{5} - 6x^{3} +2x +5] = 5x^{4} -18x^{2} + 2 + 0 </math><br><br>
+) <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>
+) <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>
-<br><br><br><br><br><br><br><br><br>
+) <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>
-<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>

2024/G2/12: Difference between revisions

Latest revision as of 16:45, 14 April 2023

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