2024/G9/12: Difference between revisions
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<math>\mathbf{Chapter 3 Section 2}</math><br> | qQ`<math>\mathbf{Chapter 3 Section 2}</math><br> | ||
<math>{\frac{d}{dx}} [c] = 0 </math> <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}} [c\cdot f(x)] = c\cdot{\frac{d}{dx}} [f(x)] </math> <br> | ||
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<math>=\frac{(2x^4+x^4+x^3+12x+6-[3x^4+3x^2-6x^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> | <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}} |
Revision as of 15:52, 30 March 2023
qQ`
<math>{\frac{d}{dx}}