2024/G2/12: Difference between revisions
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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}{ | <math>\mathbf{\color{Purple}{PracticeProblems}}</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> | ||
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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}{ | 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}{x^{2}}} + 0 - ln(-5) \cdot -5^{x} = -{\frac{3}{x^{2}}} - ln(-5) \cdot -5^{x} </math> <br><br> | ||
6) <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> |
Latest revision as of 16:45, 14 April 2023
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