Matrix multiplication

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Revision as of 10:48, 21 November 2021 by Alexandk (talk | contribs) (Created page with " Let <math> h(x) = f(x) + g(x) </math>. For all <math> x </math> near <math> a </math> we have <math> h(x) = f(x) + g(x)\approx = \left[ f(a) + f'(a) (x-a) \right] + \left[ g(...")
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Let [math]\displaystyle{  h(x) = f(x) + g(x)  }[/math]. For all [math]\displaystyle{  x  }[/math] near [math]\displaystyle{  a  }[/math] we have [math]\displaystyle{  h(x) = f(x) + g(x)\approx = \left[ f(a) + f'(a) (x-a) \right] + \left[ g(a) + g'(a) (x-a) \right] = \left[f(a) +g(a)\rigth]+ \left[f'(a) +g'(a)\right](x-a)  }[/math].  Since [math]\displaystyle{  h(a) = f(a) +g(a)  }[/math], we obtain [math]\displaystyle{  h(x)\approx h(a) + {\color{red}\left[ f'(a)+g'(a)\right]} (x-a)  }[/math]. Comparing it with [math]\displaystyle{  h(x)\approx h(a) + {\color{red} h'(a)} (x-a) }[/math] we conclude that [math]\displaystyle{  h'(a) = f'(a)+g'(a) }[/math], or (replacing [math]\displaystyle{ a }[/math] with an arbitrary variable):

[math]\displaystyle{ \boxed{(f(x)+g(x))' = f'(x)+g'(x)} }[/math]