Quiz:Product rule for differentiation

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

Practical

Corresponds to Practical:Product rule for differentiation.

General difficulty level of questions in this section: School level (unless otherwise specified).

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

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

2

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

Derivative of exponential function is exponential function, derivative of sine function is cosine function.

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

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

General difficulty level of questions in this section: College level (unless otherwise specified)

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?

	$f^{″}\cdot g+f\cdot g^{″}$
	$f^{″}\cdot g+f^{\prime }\cdot g^{\prime }+f\cdot g^{″}$
	$f^{″}\cdot g+2f^{\prime }\cdot g^{\prime }+f\cdot g^{″}$
	$f^{″}\cdot g-f^{\prime }\cdot g^{\prime }+f\cdot g^{″}$
	$f^{″}\cdot g-2f^{\prime }\cdot g^{\prime }+f\cdot g^{″}$

2 Suppose $f_1,f_2,f_3$ are everywhere differentiable functions from $\mathbb{R}$ to $\mathbb{R}$. What is the derivative $(f_1\cdot f_2\cdot f_3{)}^{\prime }$, where $f_1\cdot f_2\cdot f_3$ denotes the pointwise product of functions?

	${f_{1^{\prime }}}^{\prime }\cdot {f_{2^{\prime }}}^{\prime }\cdot {f_{3^{\prime }}}^{\prime }$
	${f_{1^{\prime }}}^{\prime }\cdot f_2\cdot f_3+f_1\cdot {f_{2^{\prime }}}^{\prime }\cdot f_3+f_1\cdot f_2\cdot {f_{3^{\prime }}}^{\prime }$
	$f_1\cdot {f_{2^{\prime }}}^{\prime }\cdot {f_{3^{\prime }}}^{\prime }+{f_{1^{\prime }}}^{\prime }\cdot f_2\cdot {f_{3^{\prime }}}^{\prime }+{f_{1^{\prime }}}^{\prime }\cdot {f_{2^{\prime }}}^{\prime }\cdot f_3$
	${f_{1^{\prime }}}^{\prime }\cdot f_2+{f_{2^{\prime }}}^{\prime }\cdot f_3+{f_{3^{\prime }}}^{\prime }\cdot f_1$
	${f_{1^{″}}}^{″}\cdot {f_{2^{\prime }}}^{\prime }\cdot f_3$

Significance

Qualitative and existential significance

Corresponds to Product rule for differentiation#Qualitative and existential significance.

General difficulty level of questions in this section: College level (unless otherwise specified).

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 $f$ and $g$ are continuous functions defined on all of $\mathbb{R}$. Suppose $A$ is the subset of $\mathbb{R}$ comprising those points where $f$ is differentiable, and $B$ is the subset of $\mathbb{R}$ comprising those points where $g$ is differentiable. Then, what can we say is definitely true about the subset of $\mathbb{R}$ comprising those points where the pointwise product of functions $f\cdot g$ is differentiable?

	It is contained in the intersection $A\cap B$
	It contains the intersection $A\cap B$
	It is contained in the union $A\cup B$
	It contains the union $A\cup B$
	None of the above

4 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

Corresponds to Product rule for differentiation#Computational results significance.

General difficulty level of questions in this section: College level (unless otherwise specified).

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

	$f\cdot g$ must be periodic
	$(f\cdot g{)}^{\prime }$ must be periodic, but $f\cdot g$ may or may not be periodic.
	$(f\cdot g{)}^{″}$ must be periodic, but $(f\cdot g{)}^{\prime }$ 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 $f\cdot g$ or 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 $f^{\prime }$ and $g^{\prime }$ are both expressible as rational functions. What can we say about $f\cdot g$ and $(f\cdot g{)}^{\prime }$ on $(0,1)$?

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