# How do you find the derivative of g(x) = -4x + 5 ?

Apr 10, 2018

$g ' \left(x\right) = - 4$

#### Explanation:

There are two ways to go about this. The first, I will be using the rules of derivatives. The second, I'll be showing by first principles.

Method 1: Rules of Derivatives

There are two rules we'll be using to solve this. The first is the derivative of a polynomial, which says:

$\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1}$

$\frac{d}{\mathrm{dx}} \left(n\right) = 0$

The highest power we have is 1, so by bringing that down, we remove the x since ${x}^{0} = 1$.

By applying both rules, we can get:

$\frac{d}{\mathrm{dx}} \left(- 4 x + 5\right) = - 4$

Method 2: First Principles

This method can be used to take the derivative of literally any function. A derivative is the instantaneous rate of change at a point, and we want the function for it. So we'll apply the rate of change formula ($\frac{r i s e}{r u n}$) and get as close to 0 as possible.

${\lim}_{h \to 0} \left(\frac{f \left(x + h\right) - f \left(x\right)}{h}\right)$

So let's plug in our function:

${\lim}_{h \to 0} \left(\frac{- 4 \left(x + h\right) + 5 - \left(- 4 x + 5\right)}{h}\right)$

Let's expand this.

${\lim}_{h \to 0} \left(\frac{- 4 x - 4 h + 5 + 4 x - 5}{h}\right)$

And let's get the like terms together:

${\lim}_{h \to 0} \left(\frac{- 4 h}{h}\right)$

And as that approaches 0, we just get

$g ' \left(x\right) = - 4$

Now, you have two methods of finding the derivative. One for polynomials, one for any other function.