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?

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

2 Suppose $f_1,f_2,f_3$ are everywhere differentiable functions from $\mathbb{R}$ to $\mathbb{R}$. What is the derivative $(f_1\circ f_2\circ f_3{)}^{\prime }$ where $\circ$ denotes the composite of two functions? In other words, $(f_1\circ f_2\circ f_3)(x):=f_1(f_2(f_3(x)))$.

	$({f_{1^{\prime }}}^{\prime }\circ f_2\circ f_3)\cdot ({f_{2^{\prime }}}^{\prime }\circ f_3)\cdot {f_{3^{\prime }}}^{\prime }$
	$({f_{1^{\prime }}}^{\prime }\cdot f_2\cdot f_3)\circ ({f_{2^{\prime }}}^{\prime }\cdot f_3)\circ {f_{3^{\prime }}}^{\prime }$
	$(f_1\circ {f_{2^{\prime }}}^{\prime }\circ {f_{3^{\prime }}}^{\prime })\cdot (f_2\circ {f_{3^{\prime }}}^{\prime })\cdot f_3$
	$(f_1\cdot {f_{2^{\prime }}}^{\prime }\cdot {f_{3^{\prime }}}^{\prime })\circ (f_2\cdot {f_{3^{\prime }}}^{\prime })\circ f_3$
	${f_{1^{\prime }}}^{\prime }\circ {f_{2^{\prime }}}^{\prime }\circ {f_{3^{\prime }}}^{\prime }$