Difference between revisions of "Derivative rules via approximations"
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<math> \boxed{(f(x)+g(x))' = f'(x)+g'(x)} </math> | <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. | + | |
| + | <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> | ||
| + | |||
| + | '''Chain rule'''. Let <math> h(x) = f(g(x)) </math>. Let <math>b=g(a)</math> and <math>y=g(x)</math> For all <math> y </math> near <math> b </math> we have | ||
| + | |||
| + | <math>f(y)\approx f(b)+f'(b)(y-b)</math> | ||
| + | |||
| + | Consequently, for all <math>x</math> near <math>a </math>: | ||
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| + | <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(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(g(x))' = f'(g(x))g'(x)} </math> | ||
Latest revision as of 13:40, 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) + {\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 [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{ 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]\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]
Product 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{ 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]\displaystyle{ h(a) = f(a)g(a) }[/math], we obtain [math]\displaystyle{ h(x)\approx h(a) + {\color{red}\left[ f'(a)g(a)+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)+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)+f(x)g'(x)} }[/math]
Chain rule. Let [math]\displaystyle{ h(x) = f(g(x)) }[/math]. Let [math]\displaystyle{ b=g(a) }[/math] and [math]\displaystyle{ y=g(x) }[/math] For all [math]\displaystyle{ y }[/math] near [math]\displaystyle{ b }[/math] we have
[math]\displaystyle{ f(y)\approx f(b)+f'(b)(y-b) }[/math]
Consequently, for all [math]\displaystyle{ x }[/math] near [math]\displaystyle{ a }[/math]:
[math]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ h(a) = f(g(a)) }[/math], we obtain [math]\displaystyle{ h(x)\approx h(a) + {\color{red}f'(g(a))g'(a)} (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'(g(a))g'(a) }[/math], or (replacing [math]\displaystyle{ a }[/math] with an arbitrary variable):
[math]\displaystyle{ \boxed{(f(g(x))' = f'(g(x))g'(x)} }[/math]