# Derivative of Sine Square Root of X

In trigonometric differentiation, most of the examples are based on the sine square roots function. We will discuss the derivative of sine square root of the $x$ function and its related examples in detail. It can be proved by the definition of differentiation.

Consider the function of the form $y = f\left( x \right) = \sin \sqrt x$

We can prove this with the help of the definition of differentiation:
$\frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{f\left( {x + \Delta x} \right) – f\left( x \right)}}{{\Delta x}}\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$

Putting the value of the function in equation (i), we get
$\frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin \sqrt {x + \Delta x} – \sin \sqrt x }}{{\Delta x}}$

Using the formula from trigonometry $\sin A – \sin B = 2\cos \left( {\frac{{A + B}}{2}} \right)\sin \left( {\frac{{A – B}}{2}} \right)$
$\frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{2\cos \left( {\frac{{\sqrt {x + \Delta x} + \sqrt x }}{2}} \right)\sin \left( {\frac{{\sqrt {x + \Delta x} – \sqrt x }}{2}} \right)}}{{\Delta x}}$

Now consider the relation
$\left( {\sqrt {x + \Delta x} + \sqrt x } \right)\left( {\sqrt {x + \Delta x} – \sqrt x } \right) = {\left( {\sqrt {x + \Delta x} } \right)^2} – {\left( {\sqrt x } \right)^2} = \Delta x$
$\begin{gathered} \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{2\cos \left( {\frac{{\sqrt {x + \Delta x} + \sqrt x }}{2}} \right)\sin \left( {\frac{{\sqrt {x + \Delta x} – \sqrt x }}{2}} \right)}}{{\left( {\sqrt {x + \Delta x} + \sqrt x } \right)\left( {\sqrt {x + \Delta x} – \sqrt x } \right)}} \\ \Rightarrow \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{2\cos \left( {\frac{{\sqrt {x + \Delta x} + \sqrt x }}{2}} \right)}}{{\left( {\sqrt {x + \Delta x} + \sqrt x } \right)}} \times \frac{{\sin \left( {\frac{{\sqrt {x + \Delta x} – \sqrt x }}{2}} \right)}}{{\left( {\sqrt {x + \Delta x} – \sqrt x } \right)}} \\ \Rightarrow \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\cos \left( {\frac{{\sqrt {x + \Delta x} + \sqrt x }}{2}} \right)}}{{\left( {\sqrt {x + \Delta x} + \sqrt x } \right)}} \times \frac{{\sin \left( {\frac{{\sqrt {x + \Delta x} – \sqrt x }}{2}} \right)}}{{\left( {\frac{{\sqrt {x + \Delta x} – \sqrt x }}{2}} \right)}} \\ \Rightarrow \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\cos \left( {\frac{{\sqrt {x + \Delta x} + \sqrt x }}{2}} \right)}}{{\left( {\sqrt {x + \Delta x} + \sqrt x } \right)}}\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin \left( {\frac{{\sqrt {x + \Delta x} – \sqrt x }}{2}} \right)}}{{\left( {\frac{{\sqrt {x + \Delta x} – \sqrt x }}{2}} \right)}} \\ \end{gathered}$

Consider $\frac{{\sqrt {x + \Delta x} – \sqrt x }}{2} = u$, as $\Delta x \to 0$, then $u \to 0$
$\begin{gathered} \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\cos \left( {\frac{{\sqrt {x + \Delta x} + \sqrt x }}{2}} \right)}}{{\left( {\sqrt {x + \Delta x} + \sqrt x } \right)}}\mathop {\lim }\limits_{u \to 0} \frac{{\sin \left( u \right)}}{{\left( u \right)}} \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{{\cos \left( {\frac{{\sqrt {x + 0} + \sqrt x }}{2}} \right)}}{{\left( {\sqrt {x + 0} + \sqrt x } \right)}}\left( 1 \right) \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{{\cos \sqrt x }}{{2\sqrt x }} \\ \end{gathered}$

Example: Find the derivative of $y = f\left( x \right) = \sin \sqrt {x – 1}$

We have the given function as
$y = \sin \sqrt {x – 1}$

Differentiating with respect to variable $x$, we get
$\frac{{dy}}{{dx}} = \frac{d}{{dx}}\sin \sqrt {x – 1}$

Using the rule, $\frac{d}{{dx}}\sin \sqrt x = \frac{{\cos \sqrt x }}{{2\sqrt x }}$, we get
$\begin{gathered} \frac{{dy}}{{dx}} = \frac{{\cos \sqrt {x – 1} }}{{2\sqrt {x – 1} }}\frac{d}{{dx}}\left( {x – 1} \right) \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{{\cos \sqrt {x – 1} }}{{2\sqrt {x – 1} }}\left( 1 \right) \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{{\cos \sqrt {x – 1} }}{{2\sqrt {x – 1} }} \\ \end{gathered}$