# What is the instantaneous rate of change of f(x)=3x+5 at x_0=1?

Aug 12, 2014

For a linear function $y = m x + b$, the rate of change, or derivative, is simply the slope m of the line. Thus, in this case the rate of change is 3 for every point along the function where the function is defined (in this case, for all real numbers $x$).

One way to prove this is to imagine what happens when we change $x$ by one unit. Let us choose a generic point P, defined as $\left({x}_{1} , {y}_{1}\right)$. If we declare ${x}_{2} = {x}_{1} + 1$, then our ${y}_{2}$ is equal to $3 \left({x}_{2}\right) + 1$ This is in turn equal to $3 \left({x}_{1} + 1\right) + 1$ which is in turn equal to $3 \left({x}_{1}\right) + 1 + 3$. From our initial equation, we see that this is equal to ${y}_{1} + 3$. Thus, for every increase of $x$ by one unit, $y$ increases 3 units.

Another is by utilizing the power rule for functions such as $f \left(x\right) = {x}^{n}$. The power rule tells us that for these functions, the derivative $\frac{d}{\mathrm{dx}} f \left(x\right) = n {x}^{n - 1}$ Since a constant function $c$ has a derivative with respect to x of 0, and the function $f \left(x\right) = x$ is the same as ${x}^{1}$, we know that $\frac{d}{\mathrm{dx}} x = 1 \left({x}^{0}\right) = 1$ Then from the constant multiple rule, we know that

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

Using those equations and the Sum Rule, which states that

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

we can apply the power rule to a function of the sort $y = m x + b$, where b, the y-intercept, is a constant.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left[m x + b\right]$

$= \frac{d}{\mathrm{dx}} m x + \frac{d}{\mathrm{dx}} b$

$= m \cdot \frac{d}{\mathrm{dx}} x + \frac{d}{\mathrm{dx}} b$

$= m \cdot 1 \left({x}^{0}\right) + 0$

$= m$