Differentiation: Sum and product rule
The product rule
We can multiply two functions #\blue{g}# and #\green{h}#. Doing so gives us the function #f(x)=\blue{g(x)}\cdot\green{h(x)}#. For #\blue{g(x)}=\blue{x^3+1}# and #\green{h(x)}=\green{x+1}#, we get #f(x)=\blue{\left(x^3+1\right)}\cdot \green{\left(x+1\right)}#.
We can calculate the derivative of a product using the product rule.
Product rule
For two differentiable functions #\blue{g}# and #\green{h}#, the product rule applies:
\[
f(x) = \blue {g(x)} \cdot \green{h(x)}\]
gives
\[f'(x)=\orange{g'(x)} \cdot \green{h(x)} + \blue{g(x)} \cdot \purple{h'(x)}
\]
Example
\[ f(x) = \blue{(2x^2+1)} \cdot \green{3x}\]
gives
\[\begin{array}{rcl}f'(x)&=& \orange{4x} \cdot \green{3x} +\blue{ (2x^2+1)} \cdot\purple{3} \\
&=&18x^2+3 \end{array}\]
We can use the following stepbystep guide to solve this.
The derivative of a product
Stepbystep 
Example 

Determine the derivative of a function which is a product of two functions: #f(x)=\blue{g(x)}\cdot \green{h(x)}#. 
#\qqquad \begin{array}{rcl}f(x)\phantom{'}&=&{(2x^2+1)}\cdot 3x\end{array}# 

Step 1 
Determine #\blue{g(x)}# and #\green{h(x)}#. 
#\qqquad\begin{array}{rcl} 
Step 2 
Determine the derivatives #\orange{g'(x)}# and #\purple{h'(x)}#. 
#\qqquad\begin{array}{rcl} 
Step 3 
Calculate the derivative of #f# using the formula: \[f'(x)= \orange{g'(x)} \cdot \green{h(x)} + \blue{g(x)} \cdot \purple{h'(x)}\] 
#\qqquad\begin{array}{rcl} f'(x)&=&\orange{4x}\cdot \green{3x}+\blue{(2x^2+1)}\cdot \purple{3} \\ 
Step 1  We determine #g(y)# and #h(y)# in such a way that #r(y)=g(y)\cdot h(y)#. #\begin{array}{rcl} g(y)&=&y^2+y+7\\ h(y)&=&y+2\end{array}# 
Step 2  We calculate the derivative #g'(y)# and #h'(y)#. #\begin{array}{rcl} g'(y)&=&\dfrac{\dd}{\dd y}\left(y^2+y+7\right)\\ &&\phantom{xxx}\blue{\text{definition derivative}}\\ &=&2\cdot y+1\\ &&\phantom{xxx}\blue{\text{sum rule, power rule and constand rule}}\end{array}# #\begin{array}{rcl} h'(y)&=&\dfrac{\dd}{\dd y}\left(y+2\right)\\ &&\phantom{xxx}\blue{\text{definition derivative}}\\ &=&1\\ &&\phantom{xxx}\blue{\text{sum rule, power rule and constand rule}}\end{array}# 
Step 3  We calculate the derivative #r'(y)#. #\begin{array}{rcl} r'\left(y\right)&=&g'(y)\cdot h(y)+g(y)\cdot h'(y)\\ &&\phantom{xxx}\blue{\text{product rule}}\\ &=&\displaystyle\left(2\cdot y+1\right)\cdot\left(y+2\right)+\left(y^2+y+7\right)\cdot\left(1\right)\\ &&\phantom{xxx}\blue{\text{entering off }f,f',g\text{ and }g'}\\ &=&\displaystyle 2\cdot y^2+5\cdot y+2+y^2+y+7\\ &&\phantom{xxx}\blue{\text{brackets expanded}}\\ &=&\displaystyle 3\cdot y^2+6\cdot y+9\\ &&\phantom{xxx}\blue{\text{simplified}} \end{array}# 
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