Question

Find the derivative of $\sqrt {ax + b}$ with respect to $x$ , using the first principles of differentiation.

Answer 1

In this question we will use the first principles of differentiation which states that when we pick up two different points on a graph, and then when move them closer to each other, we get the instantaneous rate of change which is also called the secant line.

Complete step-by-step solution:
The derivative using the first principle refers to the differential equation for the instantaneous rate of change, which is referred as:
$\dfrac{{\delta y}}{{\delta x}} = \dfrac{{f(x + h) - f(x)}}{h}$ , where $\dfrac{{\delta y}}{{\delta x}}$ is the slope of the secant line.
Now to solve this question, we will add a limit to the secant line which is true for the given differential equation.
$\mathop {\lim }\limits_{\delta x \to 0} \dfrac{{\delta y}}{{\delta x}} = \dfrac{{dy}}{{dx}} = f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$ , where $\dfrac{{dy}}{{dx}}$ is the slope of the tangent line.
Now we have to solve for the term $\sqrt {ax + b}$ therefore, on using the first principle we can write it as:
$f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sqrt {a(x + h) + b} - \sqrt {ax + b} }}{h} \to (1)$
Now we cannot directly substitute $h = 0$ since it will give us $0$ in the denominator which is a fallacy therefore, we will try to remove $h$ from the denominator.
We have the term $\dfrac{{\sqrt {a(x + h) + b} - \sqrt {ax + b} }}{h}$
On multiplying the term in the root, we get:
$\Rightarrow$ $\dfrac{{\sqrt {ax + ah + b} - \sqrt {ax + b} }}{h}$
Now on multiplying and dividing the term with $\sqrt {ax + ah + b} + \sqrt {ax + b}$ , we get the expression as:
$\Rightarrow$ $\dfrac{{\sqrt {ax + ah + b} - \sqrt {ax + b} }}{h} \times \dfrac{{\sqrt {ax + ah + b} + \sqrt {ax + b} }}{{\sqrt {ax + ah + b} + \sqrt {ax + b} }}$
Now the numerator is in the form of $(a - b)(a + b) = {a^2} - {b^2}$ therefore, we can simplify it as:
$\Rightarrow$ $\dfrac{{ax + ah + b - (ax + b)}}{{h\left( {\sqrt {ax + ah + b} + \sqrt {ax + b} } \right)}}$
On opening the bracket, we get:
$\Rightarrow$ $\dfrac{{ax + ah + b - ax - b}}{{h\left( {\sqrt {ax + ah + b} + \sqrt {ax + b} } \right)}}$
Now on simplifying, we get:
$\Rightarrow$ $\dfrac{{ah}}{{h\left( {\sqrt {ax + ah + b} + \sqrt {ax + b} } \right)}}$
Now on cancelling $h$ , we get:
$\Rightarrow$ $\dfrac{a}{{\sqrt {ax + ah + b} + \sqrt {ax + b} }}$
Therefore equation $(1)$ can be written as:
$\Rightarrow$ $f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{a}{{\sqrt {ax + ah + b} + \sqrt {ax + b} }}$
Now putting the value of the limit, we get:
$\Rightarrow$ $f'(x) = \dfrac{a}{{\sqrt {ax + a(0) + b} + \sqrt {ax + b} }}$
Which can be simplified as:
$\Rightarrow$ $f'(x) = \dfrac{a}{{\sqrt {ax + b} + \sqrt {ax + b} }}$
Which can be added and written as:
$\Rightarrow$ $f'(x) = \dfrac{a}{{2\sqrt {ax + b} }}$

Therefore the derivative is equal to $\dfrac{a}{{2\sqrt {ax + b} }}$.

Note: 1) The above question is an important result which is used in derivatives, whenever there is an expression in the form of $\sqrt {ax + b}$ its derivative is found out as $\dfrac{a}{{2\sqrt {ax + b} }}$ .
It is to be remembered that all the derivative formulas are made by using the first principles.
2) Note that you can always recheck your answer by finding the derivative of the given term, using the formulas.