Differentiation: Sum and product rule
The sum rule
When we differentiated power functions, we saw that we can factor out a constant factor. In general, we can do this when differentiating functions of the form #\orange{c}\cdot \blue{f}#.
The constant rule
For a constant #\orange{c}# and a function #\blue {f}#, the following is true:
\[ \dfrac{\dd}{\dd x}\left(\orange{c}\cdot \blue{f(x)}\right)=\orange{c}\cdot \dfrac{\dd}{\dd x}\blue{f(x)}\]
Example
\[\begin{array}{rcl}
\dfrac{\dd}{\dd x}\left(\green{\orange{3}\blue{x^3}} \right)&=&\orange{3}\cdot \dfrac{\dd}{\dd x}\blue{x^3} \\ &=& \orange{3}\cdot 3x^2 \\ &=& 9x^2\end{array}\]
We can also take the sum of two functions. The sum rule tells us what the derivative of the sum of two functions is.
For the sum of two functions #\blue{f(x)}# and #\green{g(x)}# the sum rule holds:
\[\dfrac{\dd}{\dd x}(\blue{f(x)}+\green{g(x)}) =\dfrac{\dd}{\dd x}\blue{f(x)}+\dfrac{\dd}{\dd x}\green{g(x)}\]
Example
\[\begin{array}{rcl}\dfrac{\dd}{\dd x}(\blue{x}+\green{x^2})&=&\dfrac{\dd}{\dd x}\blue{x}+\dfrac{\dd}{\dd x}\green{x^2}\\&=&1+2x\end{array}\]
#\begin{array}{rcl}
\displaystyle f'(x)&=&\displaystyle\frac{\dd}{\dd x}\left(\sqrt{5}\cdot x^2+5\cdot \sqrt{x}\right)\\
&&\phantom{xxx}\blue{\text{definition derivative}}\\
&=&\displaystyle\frac{\dd}{\dd x}\left(\sqrt{5}\cdot x^2\right)+\frac{\dd}{\dd x}\left(5\cdot \sqrt{x}\right)\\
&&\phantom{xxx}\blue{\text{sum rule }\frac{\dd}{\dd x}\left(f(x)+g(x)\right)=\frac{\dd}{\dd x}f(x)+\frac{\dd}{\dd x}g(x)}\\
&=&\displaystyle \sqrt{5}\cdot\dfrac{\dd}{\dd x}\left(x^{2}\right)+5\cdot\dfrac{\dd}{\dd x}\left(x^{{{{1}\over{2}}}}\right)\\
&&\phantom{xxx}\blue{\text{constant rule }\frac{\dd}{\dd x}\left(c\cdot f(x)\right)=c\cdot\frac{\dd}{\dd x}f(x)\text{ and }\sqrt{x}=x^{\frac{1}{2}}}\\
&=&\displaystyle \sqrt{5}\cdot\left(2\cdot x^{{1}}\right)+5\cdot\left({{1}\over{2}}\cdot x^{{-{{1}\over{2}}}}\right)\\
&&\phantom{xxx}\blue{\text{power rule }\frac{\dd}{\dd x}\left(x^n\right)=n\cdot x^{n-1}}\\
&=&\displaystyle\sqrt{5}\cdot\left(2\cdot x\right)+5\cdot\left({{1}\over{2\cdot \sqrt{x}}}\right)\\
&&\phantom{xxx}\blue{\text{terms with }x\text{ rewritten}}\\
&=&\displaystyle 2\cdot \sqrt{5}\cdot x+{{5}\over{2\cdot \sqrt{x}}}\\
&&\phantom{xxx}\blue{\text{simplified}}\\
\end{array}#
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