Product Rule

Question 1 of 4

Find the derivative

$y=(3x+4)(x-7)$

$y'=$ (6x-17)

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$\frac{dy}{dx}=\color{#9a00c7}{v}\color{#e65021}{\frac{du}{dx}}+\color{#00880A}{u}\color{#004ec4}{\frac{dv}{dx}}$

$y’=(dy)/dx$

$u’=$ $(du)/dx$

$v’=$ $(dv)/dx$

First, find the derivative of $u$ and $v$

Derivative of

$u$ :

$u$	$=$	$3x+4$
$u’$	$=$	$3$	Power Rule

Derivative of

$v$ :

$v$	$=$	$x-7$
$v’$	$=$	$1$	Power Rule

Substitute the components into the product rule

$\frac{dy}{dx}$	$=$	$\color{#9a00c7}{v}\color{#e65021}{\frac{du}{dx}}+\color{#00880A}{u}\color{#004ec4}{\frac{dv}{dx}}$

$y’$	$=$	$(\color{#9a00c7}{x-7}\cdot\color{#e65021}{3})+(\color{#00880A}{3x+4}\cdot\color{#004ec4}{1})$	Substitute known values
$y’$	$=$	$3x-21+3x+4$
$y’$	$=$	$6x-17$	Combine like terms

$y’=6x-17$

Question 2 of 4

Find the derivative

$y=x(2x-1)^6$

1. $y'=(2x+1)^5(14x-1)$
2. $y'=(2x-1)^5(14x-1)$
3. $y'=(2x-1)^5(14x+1)$
4. $y'=(2x+1)^5(14x+1)$

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Product Rule

$\frac{dy}{dx}=\color{#9a00c7}{v}\color{#e65021}{\frac{du}{dx}}+\color{#00880A}{u}\color{#004ec4}{\frac{dv}{dx}}$

$y’=(dy)/dx$

$u’=$ $(du)/dx$

$v’=$ $(dv)/dx$

First, find the derivative of $u$ and $v$

Derivative of

$u$ :

$u$	$=$	$x$
$u’$	$=$	$1$	Power Rule

Derivative of

$v$ :

$v$	$=$	$(2x-1)^6$
$v’$	$=$	$12(2x-1)^5$	Chain Rule

Substitute the components into the product rule

$\frac{dy}{dx}$	$=$	$\color{#9a00c7}{v}\color{#e65021}{\frac{du}{dx}}+\color{#00880A}{u}\color{#004ec4}{\frac{dv}{dx}}$

$y’$	$=$	$(\color{#9a00c7}{(2x-1)^6}\cdot\color{#e65021}{1})+(\color{#00880A}{x}\cdot\color{#004ec4}{12(2x-1)^5})$	Substitute known values
$y’$	$=$	$(2x-1)^6+12x(2x-1)^5$
$y’$	$=$	$(2x-1)^5[(2x-1)+12x]$	Factorize
$y’$	$=$	$(2x-1)^5(14x-1)$	Combine like terms

$y’=(2x-1)^5(14x-1)$

Question 3 of 4

Find the derivative

$y=4x(3x+1)^3$

1. $y'=4(3x+1)+36x(3x+1)^2$
2. $y'=4(3x+1)^3+36x(3x+1)$
3. $y'=4(3x+1)^3-36x(3x+1)^2$
4. $y'=4(3x+1)^3+36x(3x+1)^2$

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Product Rule

$\frac{dy}{dx}=\color{#9a00c7}{v}\color{#e65021}{\frac{du}{dx}}+\color{#00880A}{u}\color{#004ec4}{\frac{dv}{dx}}$

$y’=(dy)/dx$

$u’=$ $(du)/dx$

$v’=$ $(dv)/dx$

First, find the derivative of $u$ and $v$

Derivative of

$u$ :

$u$	$=$	$4x$
$u’$	$=$	$4$	Power Rule

Derivative of

$v$ :

$v$	$=$	$(3x+1)^3$
$v’$	$=$	$9(3x+1)^2$	Chain Rule

Substitute the components into the product rule

$\frac{dy}{dx}$	$=$	$\color{#9a00c7}{v}\color{#e65021}{\frac{du}{dx}}+\color{#00880A}{u}\color{#004ec4}{\frac{dv}{dx}}$

$y’$	$=$	$(\color{#9a00c7}{(3x+1)^3}\cdot\color{#e65021}{4})+(\color{#00880A}{4x}\cdot\color{#004ec4}{9(3x+1)^2})$	Substitute known values
$y’$	$=$	$4(3x+1)^3+36x(3x+1)^2$

$y’=4(3x+1)^3+36x(3x+1)^2$

Question 4 of 4

Find the derivative

$y=x^2(3x+2)^3$

1. $y'=(x(3x+2))(15x+4)$
2. $y'=(x(3x+2)^2)(15x-4)$
3. $y'=(x(3x+2)^2)(15x+4)$
4. $y'=((3x+2)^2)(15x+4)$

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Product Rule

$\frac{dy}{dx}=\color{#9a00c7}{v}\color{#e65021}{\frac{du}{dx}}+\color{#00880A}{u}\color{#004ec4}{\frac{dv}{dx}}$

$y’=(dy)/dx$

$u’=$ $(du)/dx$

$v’=$ $(dv)/dx$

First, find the derivative of $u$ and $v$

Derivative of

$u$ :

$u$	$=$	$x^2$
$u’$	$=$	$2x$	Power Rule

Derivative of

$v$ :

$v$	$=$	$(3x+2)^3$
$v’$	$=$	$9(3x+2)^2$	Chain Rule

Substitute the components into the product rule

$\frac{dy}{dx}$	$=$	$\color{#9a00c7}{v}\color{#e65021}{\frac{du}{dx}}+\color{#00880A}{u}\color{#004ec4}{\frac{dv}{dx}}$

$y’$	$=$	$(\color{#9a00c7}{(3x+2)^3}\cdot\color{#e65021}{2x})+(\color{#00880A}{x^2}\cdot\color{#004ec4}{9(3x+2)^2})$	Substitute known values
$y’$	$=$	$2x(3x+2)^3+9x^2(3x+2)^2$
$y’$	$=$	$x(3x+2)^2[2(3x+2)+9x]$	Factorize
$y’$	$=$	$(x(3x+2)^2)(6x+4+9x)$	Distribute
$y’$	$=$	$(x(3x+2)^2)(15x+4)$	Combine like terms

$y’=(x(3x+2)^2)(15x+4)$

Product Rule

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Well Done!

Product Rule

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Excellent!

Product Rule

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Fantastic!

Product Rule

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