Quiz:Chain rule for differentiation

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, $(f_{1}\circ f_{2}\circ f_{3})(x):=f_{1}(f_{2}(f_{3}(x)))$.

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