Product Rule

Question 1 of 4

Find the derivative

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

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

$y' =$ $y'=$ (6x-17)

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\frac{d y}{d x} = v \frac{d u}{d x} + u \frac{d v}{d x}

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

y ’ = \frac{d y}{d x}

$y’=(dy)/dx$

u ’ =

$u’=$ $\frac{d u}{d x}$ $(du)/dx$

v ’ =

$v’=$ $\frac{d v}{d x}$ $(dv)/dx$

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

Derivative of

u

$u$ :

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

Derivative of

v

$v$ :

$v$ $v$	$=$ $=$	$x - 7$ $x-7$
$v ’$ $v’$	$=$ $=$	$1$ $1$	Power Rule

Substitute the components into the product rule

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

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

y ’ = 6 x - 17

$y’=6x-17$

Question 2 of 4

Find the derivative

y = x {(2 x - 1)}^{6}

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

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

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Product Rule

\frac{d y}{d x} = v \frac{d u}{d x} + u \frac{d v}{d x}

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

y ’ = \frac{d y}{d x}

$y’=(dy)/dx$

u ’ =

$u’=$ $\frac{d u}{d x}$ $(du)/dx$

v ’ =

$v’=$ $\frac{d v}{d x}$ $(dv)/dx$

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

Derivative of

u

$u$ :

$u$ $u$	$=$ $=$	$x$ $x$
$u ’$ $u’$	$=$ $=$	$1$ $1$	Power Rule

Derivative of

v

$v$ :

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

Substitute the components into the product rule

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

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

y ’ = {(2 x - 1)}^{5} (14 x - 1)

$y’=(2x-1)^5(14x-1)$

Question 3 of 4

Find the derivative

y = 4 x {(3 x + 1)}^{3}

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

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

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Product Rule

\frac{d y}{d x} = v \frac{d u}{d x} + u \frac{d v}{d x}

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

y ’ = \frac{d y}{d x}

$y’=(dy)/dx$

u ’ =

$u’=$ $\frac{d u}{d x}$ $(du)/dx$

v ’ =

$v’=$ $\frac{d v}{d x}$ $(dv)/dx$

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

Derivative of

u

$u$ :

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

Derivative of

v

$v$ :

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

Substitute the components into the product rule

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

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

y ’ = 4 {(3 x + 1)}^{3} + 36 x {(3 x + 1)}^{2}

$y’=4(3x+1)^3+36x(3x+1)^2$

Question 4 of 4

Find the derivative

y = x^{2} {(3 x + 2)}^{3}

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

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

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Product Rule

\frac{d y}{d x} = v \frac{d u}{d x} + u \frac{d v}{d x}

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

y ’ = \frac{d y}{d x}

$y’=(dy)/dx$

u ’ =

$u’=$ $\frac{d u}{d x}$ $(du)/dx$

v ’ =

$v’=$ $\frac{d v}{d x}$ $(dv)/dx$

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

Derivative of

u

$u$ :

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

Derivative of

v

$v$ :

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

Substitute the components into the product rule

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

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

y ’ = (x {(3 x + 2)}^{2}) (15 x + 4)

$y’=(x(3x+2)^2)(15x+4)$

Product Rule

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Well Done!

Product Rule

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Nice Job!

Product Rule

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Excellent!

Product Rule

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Fantastic!

Product Rule

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