Derivative of Square Root

To find a derivative of square roots of a function. This can be proving by using derivative by definition or 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 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 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}


Differentiation 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}