Quiz:Chain rule for differentiation: Difference between revisions

Revision as of 21:39, 18 March 2024

See chain rule for differentiation and chain rule for higher derivatives for background information.

See Quiz:Differentiation rules for a quiz on all the differentiation rules together.

Formulas

1 Suppose $f$ and $g$ are both twice differentiable functions everywhere on $\mathbb {R}$ . Which of the following is the correct formula for $(f\circ g)''$ , the second derivative of the composite of two functions?

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

2 Suppose $f_{1},f_{2},f_{3}$ are everywhere differentiable functions from $\mathbb {R}$ to $\mathbb {R}$ . What is the derivative $(f_{1}\circ f_{2}\circ f_{3})'$ where $\circ$ denotes the composite of two functions? In other words, $(f_{1}\circ f_{2}\circ f_{3})(x):=f_{1}(f_{2}(f_{3}(x)))$ .

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

3 Suppose $f$ is a differentiable function from $\mathbb {R}$ to $\mathbb {R}$ and $a,b\in \mathbb {R}$ are such that $f(a)=a$ and $f'(a)=b$ . What is the value of $(f\circ f\circ \dots \circ f)'(a)$ , where $\circ$ denotes the composite of two functions and $f$ occurs $n$ times in the expression, with $n\geq 3$ ?

	$a^{n}$
	$a^{n-1}b$
	$a^{n-1}b+ab^{n-1}$
	$ab^{n-1}$
	$b^{n}$

@@ Line 31: / Line 31: @@
 - <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.
+|| 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>