Derivative of Cosine 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 cosine 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 \[y = f\left( x \right) = \cos \sqrt x \]

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
\[\frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\cos \sqrt {x + \Delta x} – \cos \sqrt x }}{{\Delta x}}\]

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

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

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

Example: Find the derivative of \[y = f\left( x \right) = \cos \sqrt {{x^2} + 1} \]

We have the given function as
\[y = \cos \sqrt {{x^2} + 1} \]

Differentiating with respect to variable $$x$$, we get
\[\frac{{dy}}{{dx}} = \frac{d}{{dx}}\cos \sqrt {{x^2} + 1} \]

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