Derivative of Sine Square Root of X

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

\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 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 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}


Differentiation 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}

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