# Derivative of a Square Root

Finding a derivative of the square roots of a function can be done by using derivative by definition or the first principle method.

Consider a function of the form $y = \sqrt x$.

First we take the increment or small change in the function.
$\begin{gathered} y + \Delta y = \sqrt {x + \Delta x} \\ \Rightarrow \Delta y = \sqrt {x + \Delta x} – y \\ \end{gathered}$

Putting the value of function $y = \sqrt x$ in the above equation, we get
$\Rightarrow \Delta y = \sqrt {x + \Delta x} – \sqrt x$

Using the rationalizing method
$\begin{gathered}\Rightarrow \Delta y = \sqrt {x + \Delta x} – \sqrt x \times \frac{{\sqrt {x + \Delta x} + \sqrt x }}{{\sqrt {x + \Delta x} + \sqrt x }} \\ \Rightarrow \Delta y = \frac{{{{\left( {\sqrt {x + \Delta x} } \right)}^2} – {{\left( {\sqrt x } \right)}^2}}}{{\sqrt {x + \Delta x} + \sqrt x }} \\ \Rightarrow \Delta y = \frac{{x + \Delta x – x}}{{\sqrt {x + \Delta x} + \sqrt x }} \\ \Rightarrow \Delta y = \frac{{\Delta x}}{{\sqrt {x + \Delta x} + \sqrt x }} \\ \end{gathered}$

Dividing both sides by $\Delta x$, we get
$\begin{gathered}\frac{{\Delta y}}{{\Delta x}} = \frac{{\Delta x}}{{\Delta x\left( {\sqrt {x + \Delta x} + \sqrt x } \right)}} \\ \frac{{\Delta y}}{{\Delta x}} = \frac{1}{{\left( {\sqrt {x + \Delta x} + \sqrt x } \right)}} \\ \end{gathered}$

Taking the limit of both sides as $\Delta x \to 0$, we have
$\begin{gathered}\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{1}{{\sqrt {x + \Delta x} + \sqrt x }} \\ \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {x + 0} + \sqrt x }} \\ \frac{{dy}}{{dx}} = \frac{1}{{2\sqrt x }}\\ \end{gathered}$

NOTE: If we take any function in the square root function, then
$\frac{{dy}}{{dx}} = \frac{1}{{2\sqrt {f\left( x \right)} }}\frac{d}{{dx}}f\left( x \right) = \frac{1}{{2\sqrt {f\left( x \right)} }}f’\left( x \right)$

Example: Find the derivative of $y = \sqrt {2{x^2} + 5}$

We have the given function as
$y = \sqrt {2{x^2} + 5}$

Differentiating with respect to variable $x$, we get
$\frac{{dy}}{{dx}} = \frac{d}{{dx}}\sqrt {2{x^2} + 5}$

Now using the formula derivative of a square root, we have
$\begin{gathered}\frac{{dy}}{{dx}} = \frac{1}{{2\sqrt {2{x^2} + 5} }}\frac{d}{{dx}}\left( {2{x^2} + 5} \right) \\ \frac{{dy}}{{dx}} = \frac{{4x}}{{2\sqrt {2{x^2} + 5} }} \\ \frac{{dy}}{{dx}} = \frac{{2x}}{{\sqrt {2{x^2} + 5} }} \\ \end{gathered}$