Difference between revisions of "Derivative rules via approximations"
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Various rules for differentiation can be derived using linear approximation. | Various rules for differentiation can be derived using linear approximation. | ||
| − | Constant multiple rule. Let <math> h(x) = k f(x) </math>, where <math>k </math> is a constant. Then for all <math>x</math> near <math>a</math> we have <math>g(x) = kf(x) \approx =k \left[ f(a) + f'(a) (x-a) \right] = kf(a) + kf'(a) (x-a).</math> Since <math> g(a) = kf(a)</math>, we obtain <math> g(x)\approx g(a) + | + | Constant multiple rule. Let <math> h(x) = k f(x) </math>, where <math>k </math> is a constant. Then for all <math>x</math> near <math>a</math> we have <math>g(x) = kf(x) \approx =k \left[ f(a) + f'(a) (x-a) \right] = kf(a) + kf'(a) (x-a).</math> Since <math> g(a) = kf(a)</math>, we obtain <math> g(x)\approx g(a) + kf'(a) (x-a) </math>. Comparing it with <math> g(x)\approx g(a) + g'(a) (x-a) </math> we conclude that <math> g'(a) = k f'(a) </math>, or (replacing <math> a </math> with an arbitrary variable): |
<math> \boxed{(kf(x))' = kf'(x)} </math> | <math> \boxed{(kf(x))' = kf'(x)} </math> | ||
Revision as of 10:28, 21 November 2021
The fact that the derivative [math]\displaystyle{ f'(a) }[/math] exists can be written as [math]\displaystyle{ f(x) \approx f(a) + f'(a) (x-a) }[/math] for all [math]\displaystyle{ x }[/math] that are near [math]\displaystyle{ a }[/math]. This is known as linear approximation (or linearization) of f at the number a. Various rules for differentiation can be derived using linear approximation.
Constant multiple rule. Let [math]\displaystyle{ h(x) = k f(x) }[/math], where [math]\displaystyle{ k }[/math] is a constant. Then for all [math]\displaystyle{ x }[/math] near [math]\displaystyle{ a }[/math] we have [math]\displaystyle{ g(x) = kf(x) \approx =k \left[ f(a) + f'(a) (x-a) \right] = kf(a) + kf'(a) (x-a). }[/math] Since [math]\displaystyle{ g(a) = kf(a) }[/math], we obtain [math]\displaystyle{ g(x)\approx g(a) + kf'(a) (x-a) }[/math]. Comparing it with [math]\displaystyle{ g(x)\approx g(a) + g'(a) (x-a) }[/math] we conclude that [math]\displaystyle{ g'(a) = k f'(a) }[/math], or (replacing [math]\displaystyle{ a }[/math] with an arbitrary variable):
[math]\displaystyle{ \boxed{(kf(x))' = kf'(x)} }[/math]
Sum/Difference rule. 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{ g(x) = f(x) \approx =k \left[ f(a) + f'(a) (x-a) \right] = kf(a) + kf'(a) (x-a) }[/math]. Since [math]\displaystyle{ g(a) = kf(a) }[/math], we obtain [math]\displaystyle{ g(x)\approx g(a) + {\color{red} kf'(a)} (x-a) }[/math]. Comparing it with [math]\displaystyle{ g(x)\approx g(a) + {\color{red} g'(a)} (x-a) }[/math] we conclude that [math]\displaystyle{ g'(a) = k f'(a) }[/math], or (replacing $a$ with an arbitrary variable):
[math]\displaystyle{ \boxed{(kf(x))' = kf'(x)} }[/math]
Product rule. Chain rule.