2024/G2/12: Difference between revisions

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1) <math>{\frac{d}{dx}} [5 + \pi] = 0+0=0 </math> <br><br>
1) <math>{\frac{d}{dx}} [5 + \pi] = 0+0=0 </math> <br><br>
2) <math>{\frac{d}{dx}} [3x] = 3 \cdot {\frac{d}{dx}} [x] = 3(1)(x)^{1-1} = 3 \cdot x^0 = 3 </math> <br><br>
2) <math>{\frac{d}{dx}} [3x] = 3 \cdot {\frac{d}{dx}} [x] = 3(1)(x)^{1-1} = 3 \cdot x^0 = 3 </math> <br><br>
3) <math>{\frac{d}{dx}} [9x^2] = 18x </math> <br><br>
3) <math>{\frac{d}{dx}} [9x^2] = 18x </math> <br><br>
4) <math>{\frac{d}{dx}} [x^3 + x^2 + 10] = 3x^2 +2x + 0 </math> <br><br>
4) <math>{\frac{d}{dx}} [x^3 + x^2 + 10] = 3x^2 +2x + 0 </math> <br><br>
5) <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>
5) <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>
A1. <math> {\sqrt[n](x^m)} = ({\sqrt[n]x})^m = x^{\frac{m}{n}} </math><br><br>
6) <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>
6) <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>
7) <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>
7) <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>
8) <math> {\frac{d}{dx}} [3x^{10}+e^{x}-5^{x}] = 30x^{9} + e^{x} - ln(5) \cdot 5^{x} </math><br><br>
8) <math> {\frac{d}{dx}} [3x^{10}+e^{x}-5^{x}] = 30x^{9} + e^{x} - ln(5) \cdot 5^{x} </math><br><br>
9) <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>
9) <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}{Practice Problems}}</math><br>
<math>\mathbf{\color{Purple}{Practice Problems}}</math><br>


1) <math> {\frac{d}{dx}} [\pi + 3x -5x^{3}] = 0+3-15x^{2} </math><br><br>
1) <math> {\frac{d}{dx}} [\pi + 3x -5x^{3}] = 0+3-15x^{2} </math><br><br>
2) <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>
2) <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>
3) <math> {\frac{d}{dx}} [x^{5} - 6x^{3} +2x +5] = 5x^{4} -18x^{2} + 2 + 0 </math><br><br>
3) <math> {\frac{d}{dx}} [x^{5} - 6x^{3} +2x +5] = 5x^{4} -18x^{2} + 2 + 0 </math><br><br>
4) <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>
 
4) <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>
 
5) <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}{2x}} + 0 -ln(-5) \cdot -5^{x} = -{\frac{3}{x^{2}} - ln(-5) \cdot -5^{x} </math><br><br>

Revision as of 21:05, 13 April 2023


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5) Failed to parse (syntax error): {\displaystyle {\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}{2x}} + 0 -ln(-5) \cdot -5^{x} = -{\frac{3}{x^{2}} - ln(-5) \cdot -5^{x} }