Quiz:Product rule for differentiation

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For a quiz that tests all the differentiation rules together, see Quiz:Differentiation rules.

Practical

1 Suppose $f$ and $g$ are both defined and differentiable at the point 1. Suppose $\!f(1)=2,g(1)=5,f'(1)=4,g'(1)=11$. What is the value of $(f\cdot g)'(1)$ where $f\cdot g$ denotes the pointwise product of functions?

	42
	44
	54
	63
	The information given is insufficient to find $(f\cdot g)'(1)$.

2

What is the derivative of the function $x\mapsto \exp(x)\sin x$? Hint for derivatives of individual functions: [SHOW MORE]

Derivative of $\exp$ is $\exp$, and derivative of $\sin$ is $\cos$.

	$x\mapsto \exp(x)(\sin x+\cos x)$
	$x\mapsto \exp(x)(\cos x-\sin x)$
	$x\mapsto \exp(x)(\sin x-\cos x)$
	$x\mapsto \exp(x)\cos x+\exp(1)\sin x$
	$x\mapsto \exp(x)\sin x+\exp(1)\cos x$

3 What is the derivative of the function $x\mapsto {\sqrt {x}}\sin(x^{2})$ for $x>0$? This question also requires use of chain rule for differentiation.

	$x\mapsto \cos(x^{2})/(2{\sqrt {x}})$
	$x\mapsto 2{\sqrt {x}}\cos(x^{2})$
	$x\mapsto 2{\sqrt {x}}(\cos(x^{2}+\sin(x^{2}))$
	$x\mapsto 2x{\sqrt {x}}\sin(x^{2})+\cos(x^{2})/(2{\sqrt {x}})$
	$x\mapsto 2x{\sqrt {x}}\cos(x^{2})+\sin(x^{2})/(2{\sqrt {x}})$

4

What is the derivative of the function $x\mapsto x\sin x\ln x$ for $x>0$? Hint for derivatives of individual functions: [SHOW MORE]

Derivative of $\sin$ is $\cos$, derivative of $\ln$ is $x\mapsto 1/x$

	$x\mapsto (\cos x)/x$
	$x\mapsto (-\cos x)/x$
	$x\mapsto \cos x\ln x+\cos x+(\sin x)/x$
	$x\mapsto \cos x\ln x-\cos x+(\sin x)/x$
	$x\mapsto \sin x\ln x+x\cos x\ln x+\sin x$

Formulas

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

	$f''\cdot g+f\cdot g''$
	$f''\cdot g+f'\cdot g'+f\cdot g''$
	$f''\cdot g+2f'\cdot g'+f\cdot g''$
	$f''\cdot g-f'\cdot g'+f\cdot g''$
	$f''\cdot g-2f'\cdot g'+f\cdot g''$

2 Suppose $f_{1},f_{2},f_{3}$ are everywhere differentiable functions from $\mathbb {R}$ to $\mathbb {R}$. What is the derivative $(f_{1}\cdot f_{2}\cdot f_{3})'$, where $f_{1}\cdot f_{2}\cdot f_{3}$ denotes the pointwise product of functions?

	$f_{1}'\cdot f_{2}'\cdot f_{3}'$
	$f_{1}'\cdot f_{2}\cdot f_{3}+f_{1}\cdot f_{2}'\cdot f_{3}+f_{1}\cdot f_{2}\cdot f_{3}'$
	$f_{1}\cdot f_{2}'\cdot f_{3}'+f_{1}'\cdot f_{2}\cdot f_{3}'+f_{1}\cdot f_{2}\cdot f_{3}'$
	$f_{1}'\cdot f_{2}+f_{2}'\cdot f_{3}+f_{3}'\cdot f_{1}$
	$f_{1}''\cdot f_{2}'\cdot f_{3}$

Significance

Qualitative and existential significance

1 Suppose $f$ and $g$ are continuous functions at $x_{0}$ and $f\cdot g$ is the pointwise product of functions. Which of the following is true (see last two options!)?

	If $f$ and $g$ are both left differentiable at $x_{0}$, then so is $f\cdot g$.
	If $f$ and $g$ are both right differentiable at $x_{0}$, then so is $f\cdot g$.
	If $f$ and $g$ are both differentiable at $x_{0}$, then so is $f\cdot g$.
	All of the above are true
	None of the above is true

2 Suppose $f$ and $g$ are continuous functions at $x_{0}$ and $f\cdot g$ is the pointwise product of functions. What is the relationship between the differentiability of $f$, $g$, and $f\cdot g$ at $x_{0}$?

	If any two of the three functions are differentiable at $x_{0}$, then so is the third.
	If $f\cdot g$ is differentiable at $x_{0}$, so are $f$ and $g$.
	If $f\cdot g$ and $f$ are differentiable at $x_{0}$, so is $g$. However, differentiability of $f$ and $g$ at $x_{0}$ does not guarantee differentiability of $f\cdot g$.
	If $f$ and $g$ are both differentiable at $x_{0}$, so is $f\cdot g$. However, differentiability of $f\cdot g$ and $f$ does not guarantee differentiability of $g$, and differentiability of $f\cdot g$ and $g$ does not guarantee differentiability of $f$.
	We cannot draw any inferences about differentiability of one of the three functions based on differentiability of the other two.

3 Suppose ${\mathcal {F}}$ is a collection of differentiable functions defined on all of $\mathbb {R}$. Further, suppose that there is a collection ${\mathcal {B}}$ of functions such that every element of ${\mathcal {F}}$ can be written as a polynomial in terms of the elements of ${\mathcal {B}}$, with constant coefficients. Suppose that the derivative of every element of ${\mathcal {B}}$ is in ${\mathcal {F}}$. Which of the following conditions are sufficient to ensure that the derivative of every element of ${\mathcal {F}}$ is in ${\mathcal {F}}$.

	It is sufficient to ensure that ${\mathcal {F}}$ is closed under addition and scalar multiplication, i.e., it forms a vector space of functions.
	It is sufficient to ensure that ${\mathcal {F}}$ is closed under multiplication, i.e., the product of any two elements of ${\mathcal {F}}$ is in ${\mathcal {F}}$.
	It is sufficient to ensure that ${\mathcal {F}}$ is closed under addition-cum-scalar multiplication (i.e., it forms a vector space), and closed under multiplication, but just having one of those conditions need not suffice.
	It is not sufficient to ensure that ${\mathcal {F}}$ is closed under addition-cum-scalar multiplication (i.e., it forms a vector space), and closed under multiplication

Computational feasibility significance

See the section #Practical.

Computational results significance

1 Suppose $f$ and $g$ are infinitely differentiable functions on all of $\mathbb {R}$ such that both $f'$ and $g'$ are periodic functions with the same period $h>0$. What can we conclude about $f\cdot g$?

	$f\cdot g$ must be periodic
	$(f\cdot g)'$ must be periodic, but $f\cdot g$ may or may not be periodic.
	$(f\cdot g)''$ must be periodic, but $(f\cdot g)'$ may or may not be periodic.
	$(f\cdot g)'''$ must be periodic, but $(f\cdot g)''$ may or may not be periodic.
	We cannot conclude from the given information whether any of the derivatives of $f\cdot g$ is periodic.

2 Suppose $f$ and $g$ are functions defined and differentiable on the open interval $(0,1)$. Suppose, further, that on $(0,1)$, the derivative functions $\!f'$ and $\!g'$ are both expressible as rational functions. What can we say about $f\cdot g$ and $(f\cdot g)'$ on $(0,1)$?

	Both $f\cdot g$ and $(f\cdot g)'$ are expressible as rational functions.
	$f\cdot g$ is expressible as a rational function, but $(f\cdot g)'$ need not be expressible as a rational function.
	$(f\cdot g)'$ is expressible as a rational function, but $f\cdot g$ need not be expressible as a rational function.
	Neither $f\cdot g$ nor $(f\cdot g)'$ need be expressible as a rational function.