Subject: Calculus

# Product Rule

## Calculus.ProductRule History

Hide minor edits - Show changes to output

June 30, 2011
by -

Changed lines 80-81 from:

{$$f'(x)=e^{2x}[ x+\frac{1}{2}$}

to:

{$$f'(x)=e^{2x}[ x+\frac{1}{2}$$}

Changed lines 85-86 from:

Solve for the derivative of {$f(x)=~~(~~2x~~)\cdot x~~^~~{2}~~$}.

to:

Solve for the derivative of {$f(x)=3y^2x^3$}.

Changed lines 90-95 from:

So,

to:

This is just direct differentiation, no more product rule involved.

{$$f'(x)=3y^2\cdot D_x x^3$$}

{$$=3y^2 \cdot \frac{1}{3}x^2$$}

{$$f'(x)=3y^2\cdot D_x x^3$$}

{$$=3y^2 \cdot \frac{1}{3}x^2$$}

Changed line 97 from:

{$$f'(x)=~~6x~~^{2}$}

to:

{$$f'(x)=y^2x^{2}$}

June 30, 2011
by -

Changed lines 65-66 from:

{$$f'(x)=2x \cdot [\sin{x}\cos{x}]+x^2\cos^2{x} - x^2\sin^2{x}$}

to:

{$$f'(x)=2x \cdot [\sin{x}\cos{x}]+x^2\cos^2{x} - x^2\sin^2{x}$$}

Changed lines 70-71 from:

Find the derivative of the function {$f$} when {$f(x)=~~(2x)~~\~~cdot x^~~{~~2~~}$}.

to:

Find the derivative of the function {$f$} when {$f(x)=\frac{1}{2}xe^{2x}$}.

Changed lines 74-80 from:

We

So,

to:

By virtue of the product rule, we proceed with direct differentiation.

{$$f'(x)=\frac{1}{2}x\cdot 2e^{2x}+\frac{1}{2}e^{2x}\cdot (1)$$}

{$$f'(x)=\frac{1}{2}x\cdot 2e^{2x}+\frac{1}{2}e^{2x}\cdot (1)$$}

Changed line 80 from:

{$$f'(x)=~~6x~~^{2}$}

to:

{$$f'(x)=e^{2x}[ x+\frac{1}{2}$}

June 30, 2011
by -

Changed lines 50-51 from:

Using Product Rule find the derivative of {$f$} when {$f(x)=~~(2x)\cdot ~~x^~~{~~2}$}.

to:

Using Product Rule find the derivative of {$f$} when {$f(x)=x^2\cdot \sin{x}\cos{x}$}.

Changed lines 55-60 from:

We ~~know~~ ~~both~~ ~~the~~ ~~derivative~~ of ~~{$2x$}~~ ~~and~~ ~~{$2x^{2}$},~~ ~~so~~ ~~let's~~ ~~differentiate~~ ~~directly~~ ~~using~~ ~~the theorem above~~.~~ ~~

So,

{$$f'(x)=~~2x~~\cdot ~~(2x) +~~x~~^~~{~~2~~}\~~cdot 2=4x^~~{~~2~~}+2x~~^~~{~~2~~}~~=6x^~~{~~2~~}$$}

So,

to:

We have a function which consists of three products, thus we may apply product rule twice.

{$$f'(x)=x^2\cdot D_x[\sin{x}\cos{x}] + 2x \cdot [\sin{x}\cos{x}]$$}

Applying again the product rule on the {$D_x[\sin{x}\cos{x}]$} term, we get

{$$f'(x)= x^2\cdot [-\sin^2{x}+\cos^2{x}] + 2x \cdot [\sin{x}\cos{x}]$$}

{$$f'(x)=x^2\cdot D_x[\sin{x}\cos{x}] + 2x \cdot [\sin{x}\cos{x}]$$}

Applying again the product rule on the {$D_x[\sin{x}\cos{x}]$} term, we get

{$$f'(x)= x^2\cdot [-\sin^2{x}+\cos^2{x}] + 2x \cdot [\sin{x}\cos{x}]$$}

Changed line 65 from:

{$$f'(x)=~~6x^~~{~~2~~}$}

to:

{$$f'(x)=2x \cdot [\sin{x}\cos{x}]+x^2\cos^2{x} - x^2\sin^2{x}$}

June 30, 2011
by -

Changed line 1 from:

'''Product Rule''' in [[differential calculus|differentiation]] applies to functions that are a multiple of each other. For example, assume we have two functions {$h$} and {$g$} and suppose further that {$f=h\cdot g$}. Then, If we are asked to find the [[derivative]] {$f$}, then we have to ~~differentiate~~ the product of the functions {$h$} and {$g$} of course. The question is, how do we do this? As for now, you maybe thinking and guessing for the answer. Some of you might even think that the derivative of {$h\cdot g$} is {$h'\cdot g'$}. Well, you have a topnotch guess. Do you know why? There is a story that even Leibniz himself, one of the Calculus' discoverer, got that same guess initially. But I have to tell you that your guess is wrong. Such error is common when it comes to this topic in derivative. Even the great Leibniz!

to:

'''Product Rule''' in [[differential calculus|differentiation]] applies to functions that are a multiple of each other. For example, assume we have two functions {$h$} and {$g$} and suppose further that {$f=h\cdot g$}. Then, If we are asked to find the [[derivative]] {$f$}, then we have to [[product rule|differentiate the product]] of the functions {$h$} and {$g$} of course. The question is, how do we do this? As for now, you maybe thinking and guessing for the answer. Some of you might even think that the derivative of {$h\cdot g$} is {$h'\cdot g'$}. Well, you have a topnotch guess. Do you know why? There is a story that even Leibniz himself, one of the Calculus' discoverer, got that same guess initially. But I have to tell you that your guess is wrong. Such error is common when it comes to this topic in derivative. Even the great Leibniz!

June 30, 2011
by -

Changed lines 14-15 from:

Find the derivative of f when {$f(x)=(2x)\cdot x^{2}$}.

to:

Find the derivative of {$f$} when {$f(x)=(2x)\cdot x^{2}$}.

Changed lines 19-20 from:

We know both the derivative of {$2x$} and {$2x^{2}$}, so let's differentiate directly using ~~according~~ ~~to~~ ~~Theorem 3~~.

to:

We know both the derivative of {$2x$} and {$2x^{2}$}, so let's differentiate directly using the theorem above.

Changed lines 23-24 from:

to:

{$$f'(x)=2x\cdot (2x) +x^{2}\cdot 2=4x^{2}+2x^{2}=6x^{2}$$}

Changed lines 27-28 from:

to:

{$$f'(x)=6x^{2}$}

Changed lines 31-101 from:

'''NEXT TOPIC''':[[Quotient Rule]]

to:

!!!Example #2

Find the derivative of the function {$f(x)=x\sin{x}$}.

(:toggle id=box2 show='Show Solution' init=hide button=1:)

>>id=box2 border='1px solid #999' padding=5px bgcolor=#edf<<

We know both the derivative of {$x$} and {$\sin{x}$}, so let's differentiate directly using the theorem above.

So,

{$$f'(x)=x\cdot \cos{x} +1\cdot \sin{x}=x\cos{x} + \sin{x}$$}

Thus,

{$$f'(x)=x\cos{x} + \sin{x}$$}

>><<

!!!Example #3

Using Product Rule find the derivative of {$f$} when {$f(x)=(2x)\cdot x^{2}$}.

(:toggle id=box3 show='Show Solution' init=hide button=1:)

>>id=box3 border='1px solid #999' padding=5px bgcolor=#edf<<

We know both the derivative of {$2x$} and {$2x^{2}$}, so let's differentiate directly using the theorem above.

So,

{$$f'(x)=2x\cdot (2x) +x^{2}\cdot 2=4x^{2}+2x^{2}=6x^{2}$$}

Thus,

{$$f'(x)=6x^{2}$}

>><<

!!!Example #4

Find the derivative of the function {$f$} when {$f(x)=(2x)\cdot x^{2}$}.

(:toggle id=box4 show='Show Solution' init=hide button=1:)

>>id=box4 border='1px solid #999' padding=5px bgcolor=#edf<<

We know both the derivative of {$2x$} and {$2x^{2}$}, so let's differentiate directly using the theorem above.

So,

{$$f'(x)=2x\cdot (2x) +x^{2}\cdot 2=4x^{2}+2x^{2}=6x^{2}$$}

Thus,

{$$f'(x)=6x^{2}$}

>><<

!!!Example #5

Solve for the derivative of {$f(x)=(2x)\cdot x^{2}$}.

(:toggle id=box5 show='Show Solution' init=hide button=1:)

>>id=box5 border='1px solid #999' padding=5px bgcolor=#edf<<

We know both the derivative of {$2x$} and {$2x^{2}$}, so let's differentiate directly using the theorem above.

So,

{$$f'(x)=2x\cdot (2x) +x^{2}\cdot 2=4x^{2}+2x^{2}=6x^{2}$$}

Thus,

{$$f'(x)=6x^{2}$}

'''NEXT TOPIC''':[[Quotient Rule]]

Find the derivative of the function {$f(x)=x\sin{x}$}.

(:toggle id=box2 show='Show Solution' init=hide button=1:)

>>id=box2 border='1px solid #999' padding=5px bgcolor=#edf<<

We know both the derivative of {$x$} and {$\sin{x}$}, so let's differentiate directly using the theorem above.

So,

{$$f'(x)=x\cdot \cos{x} +1\cdot \sin{x}=x\cos{x} + \sin{x}$$}

Thus,

{$$f'(x)=x\cos{x} + \sin{x}$$}

>><<

!!!Example #3

Using Product Rule find the derivative of {$f$} when {$f(x)=(2x)\cdot x^{2}$}.

(:toggle id=box3 show='Show Solution' init=hide button=1:)

>>id=box3 border='1px solid #999' padding=5px bgcolor=#edf<<

We know both the derivative of {$2x$} and {$2x^{2}$}, so let's differentiate directly using the theorem above.

So,

{$$f'(x)=2x\cdot (2x) +x^{2}\cdot 2=4x^{2}+2x^{2}=6x^{2}$$}

Thus,

{$$f'(x)=6x^{2}$}

>><<

!!!Example #4

Find the derivative of the function {$f$} when {$f(x)=(2x)\cdot x^{2}$}.

(:toggle id=box4 show='Show Solution' init=hide button=1:)

>>id=box4 border='1px solid #999' padding=5px bgcolor=#edf<<

We know both the derivative of {$2x$} and {$2x^{2}$}, so let's differentiate directly using the theorem above.

So,

{$$f'(x)=2x\cdot (2x) +x^{2}\cdot 2=4x^{2}+2x^{2}=6x^{2}$$}

Thus,

{$$f'(x)=6x^{2}$}

>><<

!!!Example #5

Solve for the derivative of {$f(x)=(2x)\cdot x^{2}$}.

(:toggle id=box5 show='Show Solution' init=hide button=1:)

>>id=box5 border='1px solid #999' padding=5px bgcolor=#edf<<

We know both the derivative of {$2x$} and {$2x^{2}$}, so let's differentiate directly using the theorem above.

So,

{$$f'(x)=2x\cdot (2x) +x^{2}\cdot 2=4x^{2}+2x^{2}=6x^{2}$$}

Thus,

{$$f'(x)=6x^{2}$}

'''NEXT TOPIC''':[[Quotient Rule]]

June 26, 2011
by -

Changed lines 1-2 from:

'''Product Rule''' in ~~differentiation~~ applies to functions that are a multiple of each other. For example, assume we have two functions {$h$} and {$g$} and suppose further that {$f=h\cdot g$}. Then, If we are asked to find the derivative {$f$}, then we have to differentiate the product of the functions {$h$} and {$g$} of course. The question is, how do we do this? As for now, you maybe thinking and guessing for the answer. Some of you might even think that the derivative of {$h\cdot g$} is {$h'\cdot g'$}. Well, you have a topnotch guess. Do you know why? There is a story that even Leibniz himself, one of the Calculus' discoverer, got that same guess initially. But I have to tell you that your guess is wrong. Such error is common when it comes to this topic in derivative. Even the great Leibniz!

So much for history. To make things easier, let me now show you how to find the derivative of the {$f$}.

So much for history. To make things easier, let me now show you how to find the derivative of the {$f$}.

to:

'''Product Rule''' in [[differential calculus|differentiation]] applies to functions that are a multiple of each other. For example, assume we have two functions {$h$} and {$g$} and suppose further that {$f=h\cdot g$}. Then, If we are asked to find the [[derivative]] {$f$}, then we have to differentiate the product of the functions {$h$} and {$g$} of course. The question is, how do we do this? As for now, you maybe thinking and guessing for the answer. Some of you might even think that the derivative of {$h\cdot g$} is {$h'\cdot g'$}. Well, you have a topnotch guess. Do you know why? There is a story that even Leibniz himself, one of the Calculus' discoverer, got that same guess initially. But I have to tell you that your guess is wrong. Such error is common when it comes to this topic in derivative. Even the great Leibniz!

So much for history. To make things easier, let me now show you how to find the [[derivative]] of the {$f$}.

So much for history. To make things easier, let me now show you how to find the [[derivative]] of the {$f$}.

June 26, 2011
by -

Added lines 16-18:

(:toggle id=box1 show='Show Solution' init=hide button=1:)

>>id=box1 border='1px solid #999' padding=5px bgcolor=#edf<<

>>id=box1 border='1px solid #999' padding=5px bgcolor=#edf<<

Added lines 28-29:

>><<

June 26, 2011
by -

Changed line 13 from:

!!!~~Example~~

to:

!!!Example #1

Added lines 25-26:

'''NEXT TOPIC''':[[Quotient Rule]]

November 27, 2010
by -

Changed line 24 from:

%cframe% '+{$f'(x)=6x^{~~3~~}$}+'

to:

%cframe% '+{$f'(x)=6x^{2}$}+'

November 27, 2010
by -

Changed lines 16-19 from:

We know both the derivative of {$2x$} and {$2x^{2}$}, so let's differentiate directly using according to Theorem 3. So,

%cframe~~%~~ '+{$f'(x)=2x\cdot (2x) +x^{2}\cdot 2=4x^{2}+2x^{2}=6x^{2}$}+'~~. ~~

%cframe

to:

We know both the derivative of {$2x$} and {$2x^{2}$}, so let's differentiate directly using according to Theorem 3.

So,

%cframe width=320px% '+{$f'(x)=2x\cdot (2x) +x^{2}\cdot 2=4x^{2}+2x^{2}=6x^{2}$}+'

So,

%cframe width=320px% '+{$f'(x)=2x\cdot (2x) +x^{2}\cdot 2=4x^{2}+2x^{2}=6x^{2}$}+'

Changed line 24 from:

%cframe% '+{$f'(x)=6x^{3}$}+'~~.~~

to:

%cframe% '+{$f'(x)=6x^{3}$}+'

November 27, 2010
by -

Changed lines 16-22 from:

We know both the derivative of {$2x$} and {$2x^{2}$}, so let's differentiate directly using according to Theorem 3. So, {$f'(x)=2x\cdot (2x) +x^{2}\cdot 2=4x^{2}+2x^{2}=6x^{2}$}. Thus, {$f'(x)=6x^{3}$}.

to:

We know both the derivative of {$2x$} and {$2x^{2}$}, so let's differentiate directly using according to Theorem 3. So,

%cframe% '+{$f'(x)=2x\cdot (2x) +x^{2}\cdot 2=4x^{2}+2x^{2}=6x^{2}$}+'.

Thus,

%cframe% '+{$f'(x)=6x^{3}$}+'.

%cframe% '+{$f'(x)=2x\cdot (2x) +x^{2}\cdot 2=4x^{2}+2x^{2}=6x^{2}$}+'.

Thus,

%cframe% '+{$f'(x)=6x^{3}$}+'.

November 11, 2010
by -

Changed lines 4-5 from:

to:

!!Theorem (Product Rule)

(:table border=3 cellpadding=3 cellspacing=0 align=center:)

(:cellnr:)

%center%'+If {$h$} and {$g$} are differentiable at {$x$}, and {$f(x)=h(x)\cdot g(x)$},\\

then {$f'(x)=g(x)\cdot h'(x) + h(x)\cdot g'(x)$}+'.

(:tableend:)

(:table border=3 cellpadding=3 cellspacing=0 align=center:)

(:cellnr:)

%center%'+If {$h$} and {$g$} are differentiable at {$x$}, and {$f(x)=h(x)\cdot g(x)$},\\

then {$f'(x)=g(x)\cdot h'(x) + h(x)\cdot g'(x)$}+'.

(:tableend:)

November 09, 2010
by -

Changed lines 1-2 from:

'''Product Rule'''

If {$h$} and {$g$} are differentiable at {$x$}, and {$f(x)=h(x)\cdot g(x)$}, then {$f'(x)=g(x)\cdot h'(x) + h(x)\cdot g'(x)$}

If {$h$} and {$g$} are differentiable at {$x$}, and {$f(x)=h(x)\cdot g(x)$}, then {$f'(x)=g(x)\cdot h'(x) + h(x)\cdot g'(x)$}

to:

'''Product Rule''' in differentiation applies to functions that are a multiple of each other. For example, assume we have two functions {$h$} and {$g$} and suppose further that {$f=h\cdot g$}. Then, If we are asked to find the derivative {$f$}, then we have to differentiate the product of the functions {$h$} and {$g$} of course. The question is, how do we do this? As for now, you maybe thinking and guessing for the answer. Some of you might even think that the derivative of {$h\cdot g$} is {$h'\cdot g'$}. Well, you have a topnotch guess. Do you know why? There is a story that even Leibniz himself, one of the Calculus' discoverer, got that same guess initially. But I have to tell you that your guess is wrong. Such error is common when it comes to this topic in derivative. Even the great Leibniz!

So much for history. To make things easier, let me now show you how to find the derivative of the {$f$}.

%cframe align=cente%If {$h$} and {$g$} are differentiable at {$x$}, and {$f(x)=h(x)\cdot g(x)$}, then {$f'(x)=g(x)\cdot h'(x) + h(x)\cdot g'(x)$}

So much for history. To make things easier, let me now show you how to find the derivative of the {$f$}.

%cframe align=cente%If {$h$} and {$g$} are differentiable at {$x$}, and {$f(x)=h(x)\cdot g(x)$}, then {$f'(x)=g(x)\cdot h'(x) + h(x)\cdot g'(x)$}

November 08, 2010
by -

November 08, 2010
by -

Added lines 1-6:

If {$h$} and {$g$} are differentiable at {$x$}, and {$f(x)=h(x)\cdot g(x)$}, then {$f'(x)=g(x)\cdot h'(x) + h(x)\cdot g'(x)$}

!!!Example

Find the derivative of f when {$f(x)=(2x)\cdot x^{2}$}.

We know both the derivative of {$2x$} and {$2x^{2}$}, so let's differentiate directly using according to Theorem 3. So, {$f'(x)=2x\cdot (2x) +x^{2}\cdot 2=4x^{2}+2x^{2}=6x^{2}$}. Thus, {$f'(x)=6x^{3}$}.

!!!Example

Find the derivative of f when {$f(x)=(2x)\cdot x^{2}$}.

We know both the derivative of {$2x$} and {$2x^{2}$}, so let's differentiate directly using according to Theorem 3. So, {$f'(x)=2x\cdot (2x) +x^{2}\cdot 2=4x^{2}+2x^{2}=6x^{2}$}. Thus, {$f'(x)=6x^{3}$}.