Quiz:Chain rule for differentiation: Difference between revisions

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 - <math>(f_1 \cdot f_2' \cdot f_3') \circ (f_2 \cdot f_3') \circ f_3</math>
 - <math>f_1' \circ f_2' \circ f_3'</math>
+{ Suppose <math>f</math> is a differentiable function from <math>\R</math> to <math>\R</math> and <math>a, b \in \R</math> are such that <math>f(a) = a</math> and <math>f'(a) = b</math>. What is the value of <math>(f \circ f \circ \dots \circ f)'(a)</math>, where <math>\circ</math> denotes the [[composite of two functions]] and <math>f</math> occurs <math>n</math> times in the expression, with <math>n \ge 3</math>?
+|type="()"}
+- <math>a^n</math>
+- <math>a^{n - 1}b</math>
+- <math>a^{n - 1}b + ab^{n - 1}</math>
+- <math>ab^{n - 1}</math>
++ <math>b^n</math>
+|| The chain rule gives the derivative as a product of <math>n</math> terms, each of which is of the form <math>f'<math> applied to <math>k</math> iterates of <math>f</math>, with <math>k</math> varying from <math>0</math> to <math>n - 1</math>. Evaluating at <math>a</math> and using <math>f'(a) = a</math>, each term simplifies to <math>f'(a)</math> and hence to <math>b</math>. As there are <math>n</math> such terms, the product is <math>b^n</math>. Note that <math>n \ge 3</math> is not necessary (this reasoning works for <math>n = 1</math> and <math>n = 2</math> as well). That condition was added primarily to dissuade people from using <math>n = 1</math> or <math>n = 2</math> to figure out the correct answer by a process of elimination.
 </quiz>

	$(f^{″} \circ g) \cdot g^{″}$
	$(f^{″} \circ g) \cdot (f^{'} \circ g^{'}) \cdot g^{″}$
	$(f^{″} \circ g) \cdot (f^{'} \circ g^{'}) \cdot (f \circ g^{″})$
	$(f^{″} \circ g) \cdot (g^{'})^{2} + (f^{'} \circ g) \cdot g^{″}$
	$(f^{'} \circ g^{'}) \cdot (f \circ g) + (f^{″} \circ g^{″})$

	$({f_{1^{'}}}^{'} \circ f_{2} \circ f_{3}) \cdot ({f_{2^{'}}}^{'} \circ f_{3}) \cdot {f_{3^{'}}}^{'}$
	$({f_{1^{'}}}^{'} \cdot f_{2} \cdot f_{3}) \circ ({f_{2^{'}}}^{'} \cdot f_{3}) \circ {f_{3^{'}}}^{'}$
	$(f_{1} \circ {f_{2^{'}}}^{'} \circ {f_{3^{'}}}^{'}) \cdot (f_{2} \circ {f_{3^{'}}}^{'}) \cdot f_{3}$
	$(f_{1} \cdot {f_{2^{'}}}^{'} \cdot {f_{3^{'}}}^{'}) \circ (f_{2} \cdot {f_{3^{'}}}^{'}) \circ f_{3}$
	${f_{1^{'}}}^{'} \circ {f_{2^{'}}}^{'} \circ {f_{3^{'}}}^{'}$

Revision as of 05:37, 5 December 2023