Rule for sums, differences, and constant multiples
If $f\prime(x)$ and $g\prime(x)$ exist, and $c$ is a constant, then
a. $\displaystyle [f(x) \pm g(x)]\prime = f'(x) \pm g\prime(x)$
b. $\displaystyle [c\,f(x)]\prime = c\,f\prime(x) $
In differential notation, these rules can be written as follows:
a. $\displaystyle \frac{d}{dx}[f(x) \pm g(x)] = \frac{d}{dx}[f(x)] \pm \frac{d}{dx}[g(x)]$
b. $\displaystyle \frac{d}{dx}[c\,f(x)] = c\frac{d}{dx}[f(x)] $
In words
The derivative of a sum is the sum of the derivatives, and the derivative of a difference is the difference of the derivatives.
In other words, to find the derivative of a sum (or difference) of several functions, just find the derivative of each function, and add (or subtract) the answers.
The derivative of $c$ times a function is $c$ times the derivative of the function.
In other words, to find the derivative of a constant times a function, just find the derivative of the function, and multiply by the constant.
#[Examples][Ejemplos]#
1. \t $\displaystyle \frac{d}{dx}[1+x^3]$ \t ${}= 0 + 3x^2$ \gap[10] \t
\\ \t \t ${}= 3x^2$
\\
\\ 2. \t $\displaystyle \frac{d}{dx}[x^2-x^3+x^5]$ \t ${}= 2x-3x^2+5x^4$ \gap[10] \t
\\
3. \t $\displaystyle \frac{d}{dx}[4x^3]$ \t $\displaystyle {}= 4\frac{d}{dx}[x^3]$ \gap[10] \t
\\ \t \t ${}= 4(3x^2)$
\\ \t \t ${}= 12x^2$ \t
\\
4. \t $\displaystyle \frac{d}{dx}[12]$ \t $\displaystyle {}= \frac{d}{dx}[12 \cdot 1]$ \gap[10] \t
\\ \t \t $\displaystyle {}= 12\frac{d}{dx}[1]$ \gap[10] \t
\\ \t \t ${}= 12(0)$ \t
\\ \t \t ${}= 0$ \t
\\
5. \t $\displaystyle \frac{d}{dx}\left[\frac{4}{x}\right]$ \t $\displaystyle {}= \frac{d}{dx}\left[4 \cdot \frac{1}{x}\right]$ \gap[10] \t
\\ \t \t $\displaystyle {}= 4\frac{d}{dx}\left[\frac{1}{x}\right]$ \t
\\ \t \t $\displaystyle {}= 4\left[-\frac{1}{x^2}\right]$ \t
\\ \t \t $\displaystyle {}= -\frac{4}{x^2}$
\\
\\ 6. \t $\displaystyle \frac{d}{dx}[5x^3-4x+7]$ \t ${}= 5(3x^2) - 4(1) + 7(0)$ \gap[10] \t
\\ \t \t $\displaystyle {}= 15x^2-4$