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

Jun 9, 2015

Assuming $a$ and $b$ are constants, the answer is

$f ' \left(x\right) = \frac{a}{2 \sqrt{a x + b}}$.

#### Explanation:

The first principle of differentiation states that if you pick two points of a graph (the average rate of change or the secant line) and move them closer to each other, you will get the instantaneous rate of change (or the tangent line where it is only one point). So we can pick one point as $x$ and the other $x + h$, where $h = \delta x$ or the change from the initial $x$ to the next point. See http://wwwf.imperial.ac.uk/metric/metric_public/differentiation/ for more information.

The partial differentiation equation for a secant line is referred to as:

(1) $\frac{\delta y}{\delta x} = \frac{f \left(x + h\right) - f \left(x\right)}{x + h - x} = \frac{f \left(x + h\right) - f \left(x\right)}{h}$,
where $\frac{\delta y}{\delta x}$ is the slope of a secant line.

But, in this problem, we have to include a limit definition for the tangent line, which is the true differentiation equation:

(2) ${\lim}_{\delta x \to 0} \frac{\delta y}{\delta x} = \frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right) = {\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$,
where $\frac{\mathrm{dy}}{\mathrm{dx}}$ is the slope of a tangent line.

Now we have the information to solve for the differentiation of $f \left(x\right) = \sqrt{a x + b}$ using the first principle. First, insert the variables for the function using Equation 2:

(3) $f ' \left(x\right) = {\lim}_{h \to 0} \frac{\sqrt{a \left(x + h\right) + b} - \sqrt{a x + b}}{h}$

We can't make the conclusion yet since inserting $h = 0$ for the limit gives us a divide by zero error. So we need to somehow factor out the square roots to the bottom as such with a clever one:

$\frac{\sqrt{a \left(x + h\right) + b} - \sqrt{a x + b}}{h}$

$= \frac{\sqrt{a x + a h + b} - \sqrt{a x + b}}{h} \cdot \frac{\sqrt{a x + a h + b} + \sqrt{a x + b}}{\sqrt{a x + a h + b} + \sqrt{a x + b}}$

$= \frac{\cancel{a x} + a h + \cancel{b} - \left(\cancel{a x} + \cancel{b}\right)}{h \left(\sqrt{a x + a h + b} + \sqrt{a x + b}\right)}$

=(acancel(h))/(cancel(h)(sqrt(ax+ah+b)+sqrt(ax+b))

=a/(sqrt(a(x+h)+b)+sqrt(ax+b)

Now we substitute it back to Eq. 3 and solve for the limit differentiation:

f'(x)=lim_(hto0)a/(sqrt(a(x+h)+b)+sqrt(ax+b)

$f ' \left(x\right) = \frac{a}{\sqrt{a \left(x + 0\right) + b} + \sqrt{a x + b}} = \frac{a}{2 \sqrt{a x + b}}$

Later on in calculus, you will learn the shortcuts to solve for these kinds of problems, such as the Power Rule and the Chain Rule. Both were used to solve for the answer, which correctly correlates to our first principle answer.