The sum rule

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

$\begin{array}{rcl} \dfrac{\dd}{\dd x}\left(\green{\orange{3}\blue{x^3}} \right)&=&\orange{3}\cdot \dfrac{\dd}{\dd x}\blue{x^3} \\ &=& \orange{3}\cdot 3x^2 \\ &=& 9x^2\end{array}$

The derivative of a constant times a function $c\cdot f(x)$ is:

$\dfrac{\dd}{\dd x}\left( c\cdot f(x)\right)= \dfrac{c\cdot f(x+h) - c\cdot f(x)}{h} = c \cdot \dfrac{f(x+h) - f(x)}{h}= c \cdot \dfrac{\dd}{\dd x} f(x)$

For $h \to 0$ .

We can also take the sum of two functions. The sum rule tells us what the derivative of the sum of two functions is.

The sum rule

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

$\begin{array}{rcl}\dfrac{\dd}{\dd x}(\blue{x}+\green{x^2})&=&\dfrac{\dd}{\dd x}\blue{x}+\dfrac{\dd}{\dd x}\green{x^2}\\&=&1+2x\end{array}$

The derivative of the sum of two functions $f(x)+g(x)$ is:

$\begin{array}{rcl} \dfrac{\dd}{\dd x}(f(x)+g(x)) &=& \dfrac{f(x+h)+g(x+h)-(f(x)+g(x))}{h}\\&=&\dfrac{f(x+h)-f(x)}{h} + \dfrac{g(x+h)-g(x)}{h}\\&=&\dfrac{\dd}{\dd x}f(x)+\dfrac{\dd}{\dd x}g(x)\end{array}$

for $h \to 0$ .

Calculate the derivative $f'(x)$ of the function $f(x)=$ $\sqrt{6}\cdot x^4+6\cdot \sqrt{x}$ .

$f'(x)=$ $4\cdot \sqrt{6}\cdot x^3+{{3}\over{\sqrt{x}}}$

$\begin{array}{rcl} \displaystyle f'(x)&=&\displaystyle\frac{\dd}{\dd x}\left(\sqrt{6}\cdot x^4+6\cdot \sqrt{x}\right)\\ &&\phantom{xxx}\blue{\text{definition derivative}}\\ &=&\displaystyle\frac{\dd}{\dd x}\left(\sqrt{6}\cdot x^4\right)+\frac{\dd}{\dd x}\left(6\cdot \sqrt{x}\right)\\ &&\phantom{xxx}\blue{\text{sum rule }\frac{\dd}{\dd x}\left(f(x)+g(x)\right)=\frac{\dd}{\dd x}f(x)+\frac{\dd}{\dd x}g(x)}\\ &=&\displaystyle \sqrt{6}\cdot\dfrac{\dd}{\dd x}\left(x^{4}\right)+6\cdot\dfrac{\dd}{\dd x}\left(x^{{{{1}\over{2}}}}\right)\\ &&\phantom{xxx}\blue{\text{constant rule }\frac{\dd}{\dd x}\left(c\cdot f(x)\right)=c\cdot\frac{\dd}{\dd x}f(x)\text{ and }\sqrt{x}=x^{\frac{1}{2}}}\\ &=&\displaystyle \sqrt{6}\cdot\left(4\cdot x^{{3}}\right)+6\cdot\left({{1}\over{2}}\cdot x^{{-{{1}\over{2}}}}\right)\\ &&\phantom{xxx}\blue{\text{power rule }\frac{\dd}{\dd x}\left(x^n\right)=n\cdot x^{n-1}}\\ &=&\displaystyle\sqrt{6}\cdot\left(4\cdot x^3\right)+6\cdot\left({{1}\over{2\cdot \sqrt{x}}}\right)\\ &&\phantom{xxx}\blue{\text{terms with }x\text{ rewritten}}\\ &=&\displaystyle 4\cdot \sqrt{6}\cdot x^3+{{3}\over{\sqrt{x}}}\\ &&\phantom{xxx}\blue{\text{simplified}}\\ \end{array}$

New example