Derivative rules via approximations - Revision history

Alexandk at 17:40, 21 November 2021

2021-11-21T17:40:15Z

2021-11-21T17:38:50Z

2021-11-21T17:38:27Z

2021-11-21T17:37:24Z

2021-11-21T17:31:04Z

2021-11-21T17:29:26Z

2021-11-21T16:34:40Z

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 <math>f(g(a)) + f'(g(a)) (g'(a)(x-a)) = f(g(a)) + {\color{red} f'(g(a))g'(a)} (x-a)</math>.
-Since <math> h(a) = f(a)g(a) </math>, we obtain <math> h(x)\approx h(a) + {\color{red}\left[ f'(a)g(a)+f(a)g'(a)\right]} (x-a) </math>. Comparing it with <math> h(x)\approx h(a) + {\color{red} h'(a)} (x-a)</math> we conclude that <math> h'(a) = f'(a)g(a)+f(a)g'(a)</math>, or (replacing <math>a</math> with an arbitrary variable):
+Since <math> h(a) = f(g(a))</math>, we obtain <math> h(x)\approx h(a) + {\color{red}f'(g(a))g'(a)} (x-a) </math>. Comparing it with <math> h(x)\approx h(a) + {\color{red} h'(a)} (x-a)</math> we conclude that <math> h'(a) = f'(g(a))g'(a)</math>, or (replacing <math>a</math> with an arbitrary variable):
-<math> \boxed{(f(x)g(x))' = f'(x)g(x)+f(x)g'(x)} </math>
+<math> \boxed{(f(g(x))' = f'(g(x))g'(x)} </math>

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 Consequently, for all <math>x</math> near <math>a </math>:
-<math>h(x)= f(g(x)) = f(y)\approx f(b) + f'(b) (y-b) = f(g(a)) + f'(g(a))(g(x) - b)\approx f(g(a)) +f'(g(a))\left[ (g(a)+ g'(a)(x-a)\right] - g(a) ) = </math> <math>f(g(a)) + f'(g(a)) (g'(a)(x-a)) = f(g(a)) + {\color{red} f'(g(a))g'(a)} (x-a)</math>.
+<math>h(x)= f(g(x)) = f(y)\approx f(b) + f'(b) (y-b) = f(g(a)) + f'(g(a))(g(x) - b)\approx f(g(a)) +f'(g(a))\left[ (g(a)+ g'(a)(x-a)\right] - g(a) ) = </math>
 Since <math> h(a) = f(a)g(a) </math>, we obtain <math> h(x)\approx h(a) + {\color{red}\left[ f'(a)g(a)+f(a)g'(a)\right]} (x-a) </math>. Comparing it with <math> h(x)\approx h(a) + {\color{red} h'(a)} (x-a)</math> we conclude that <math> h'(a) = f'(a)g(a)+f(a)g'(a)</math>, or (replacing <math>a</math> with an arbitrary variable):
 <math> \boxed{(f(x)g(x))' = f'(x)g(x)+f(x)g'(x)} </math>

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 Consequently, for all <math>x</math> near <math>a </math>:
-<math>h(x)= f(g(x)) = f(y)\approx f(b) + f'(b) (y-b) = f(g(a)) + f'(g(a))(g(x) - b)\approx f(g(a)) +f'(g(a))\left[ (g(a)+ g'(a)(x-a)\right] - g(a) ) = f(g(a)) + f'(g(a)) (g'(a)(x-a)) = f(g(a)) + {\color{red} f'(g(a))g'(a)} (x-a)</math>.
+<math>h(x)= f(g(x)) = f(y)\approx f(b) + f'(b) (y-b) = f(g(a)) + f'(g(a))(g(x) - b)\approx f(g(a)) +f'(g(a))\left[ (g(a)+ g'(a)(x-a)\right] - g(a) ) = </math> <math>f(g(a)) + f'(g(a)) (g'(a)(x-a)) = f(g(a)) + {\color{red} f'(g(a))g'(a)} (x-a)</math>.
 Since <math> h(a) = f(a)g(a) </math>, we obtain <math> h(x)\approx h(a) + {\color{red}\left[ f'(a)g(a)+f(a)g'(a)\right]} (x-a) </math>. Comparing it with <math> h(x)\approx h(a) + {\color{red} h'(a)} (x-a)</math> we conclude that <math> h'(a) = f'(a)g(a)+f(a)g'(a)</math>, or (replacing <math>a</math> with an arbitrary variable):
 <math> \boxed{(f(x)g(x))' = f'(x)g(x)+f(x)g'(x)} </math>

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 <math> h(x) = f(x) g(x)\approx \left[ f(a) + f'(a) (x-a) \right] \cdot \left[ g(a) + g'(a) (x-a) \right] = f(a)\cdot g(a)+ {\color{red} \left[f'(a)g(a) +f(a)g'(a)\right]}(x-a) + {\color{green} f'(a)g'(a)(x-a)^2}\</math>.
 Since <math> h(a) = f(a)g(a) </math>, we obtain <math> h(x)\approx h(a) + {\color{red}\left[ f'(a)g(a)+f(a)g'(a)\right]} (x-a) </math>. Comparing it with <math> h(x)\approx h(a) + {\color{red} h'(a)} (x-a)</math> we conclude that <math> h'(a) = f'(a)g(a)+f(a)g'(a)</math>, or (replacing <math>a</math> with an arbitrary variable):
 <math> \boxed{(f(x)g(x))' = f'(x)g(x)+f(x)g'(x)} </math>

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 '''Product rule'''. 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] \cdot \left[ g(a) + g'(a) (x-a) \right] = \f(a)\cdotg(a)+ {\color{red} \left[f'(a)g(a) +g'(a)f(a)\right]}(x-a) + {\color{green} f'(a)g'(a)(x-a)^2}\</math>.
+<math> h(x) = f(x) g(x)\approx \left[ f(a) + f'(a) (x-a) \right] \cdot \left[ g(a) + g'(a) (x-a) \right] = f(a)\cdot g(a)+ {\color{red} \left[f'(a)g(a) +f(a)g'(a)\right]}(x-a) + {\color{green} f'(a)g'(a)(x-a)^2}\</math>.
-Since <math> h(a) = f(a) +g(a) </math>, we obtain <math> h(x)\approx h(a) + {\color{red}\left[ f'(a)+g'(a)\right]} (x-a) </math>. Comparing it with <math> h(x)\approx h(a) + {\color{red} h'(a)} (x-a)</math> we conclude that <math> h'(a) = f'(a)+g'(a)</math>, or (replacing <math>a</math> with an arbitrary variable):
+Since <math> h(a) = f(a)g(a) </math>, we obtain <math> h(x)\approx h(a) + {\color{red}\left[ f'(a)g(a)+f(a)g'(a)\right]} (x-a) </math>. Comparing it with <math> h(x)\approx h(a) + {\color{red} h'(a)} (x-a)</math> we conclude that <math> h'(a) = f'(a)g(a)+f(a)g'(a)</math>, or (replacing <math>a</math> with an arbitrary variable):
-<math> \boxed{(f(x)+g(x))' = f'(x)+g'(x)} </math>
+<math> \boxed{(f(x)g(x))' = f'(x)g(x)+f(x)g'(x)} </math>

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 <math> \boxed{(f(x)+g(x))' = f'(x)+g'(x)} </math>
-Product rule.
+'''Product rule'''. Let <math> h(x) = f(x)g(x) </math>. For all <math> x </math> near <math> a </math> we have
-Chain rule.

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 <math> \boxed{(kf(x))' = kf'(x)} </math>
-'''Sum/Difference rule'''. 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(a) + g'(a) (x-a) \right] = \left[f(a) +g(a)\right]+ \left[f'(a) +g'(a)\right](x-a) </math>. Since <math> h(a) = f(a) +g(a) </math>, we obtain <math> h(x)\approx h(a) + {\color{red}\left[ f'(a)+g'(a)\right]} (x-a) </math>. Comparing it with <math> h(x)\approx h(a) + {\color{red} h'(a)} (x-a)</math> we conclude that <math> h'(a) = f'(a)+g'(a)</math>, or (replacing <math>a</math> with an arbitrary variable):
+'''Sum/Difference rule'''. Let <math> h(x) = f(x) + g(x) </math>. For all <math> x </math> near <math> a </math> we have
 <math> \boxed{(f(x)+g(x))' = f'(x)+g'(x)} </math>