# How do you find the derivative of f(x)=ax+b?

Sep 25, 2017

${f}^{'} \left(x\right) = a$

#### Explanation:

$f \left(x\right) = a x + b$

Take derivative on both sides:

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \frac{d}{\mathrm{dx}} \left(a x + b\right)$

Apply the sum/difference rule for derivative which is stated as:

$\frac{d}{\mathrm{dx}} \left(f + g\right) = \frac{d}{\mathrm{dx}} \left(f\right) + \frac{d}{\mathrm{dx}} \left(g\right)$

So that we will have:

${f}^{'} \left(x\right) = \frac{d}{\mathrm{dx}} \left(a x\right) + \frac{d}{\mathrm{dx}} \left(b\right)$

Remember the derivative of a constant is zero, so that we will have:

${f}^{'} \left(x\right) = \frac{d}{\mathrm{dx}} \left(a x\right) + 0$

Take the constant out by applying $\frac{d}{\mathrm{dx}} \left(a \cdot f\right) = a \cdot \frac{d}{\mathrm{dx}} \left(f\right)$

${f}^{'} \left(x\right) = a \cdot \frac{d}{\mathrm{dx}} \left(x\right)$

Apply the common derivative rule $\frac{d}{\mathrm{dx}} \left(x\right) = 1$

${f}^{'} \left(x\right) = a \cdot 1$

${f}^{'} \left(x\right) = a$