# Derivative of Cosine

We shall prove the formula for the derivative of the cosine function by using definition or the first principle method.

Let us suppose that the function is of the form $y = f\left( x \right) = \cos x$.

First we take the increment or small change in the function:
$\begin{gathered} y + \Delta y = \cos \left( {x + \Delta x} \right) \\ \Delta y = \cos \left( {x + \Delta x} \right) – y \\ \end{gathered}$

Putting the value of function $y = \cos x$ in the above equation, we get
$\Delta y = \cos \left( {x + \Delta x} \right) – \cos x\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$

Using the formula from trigonometry, we have
$\cos A – \cos B = – 2\sin \left( {\frac{{A + B}}{2}} \right)\sin \left( {\frac{{A – B}}{2}} \right)$

Using this formula in equation (i), we get
$\begin{gathered} \Delta y = – 2\sin \left( {\frac{{x + \Delta x + x}}{2}} \right)\sin \left( {\frac{{x + \Delta x – x}}{2}} \right) \\ \Delta y = – 2\sin \left( {\frac{{2x + \Delta x}}{2}} \right)\sin \left( {\frac{{\Delta x}}{2}} \right) \\ \end{gathered}$

Dividing both sides by $\Delta x$, we get
$\begin{gathered} \Rightarrow \frac{{\Delta y}}{{\Delta x}} = \frac{{ – 2\sin \left( {\frac{{2x + \Delta x}}{2}} \right)\sin \left( {\frac{{\Delta x}}{2}} \right)}}{{\Delta x}} \\ \Rightarrow \frac{{\Delta y}}{{\Delta x}} = – \frac{{\sin \left( {\frac{{2x + \Delta x}}{2}} \right)\sin \left( {\frac{{\Delta x}}{2}} \right)}}{{\frac{{\Delta x}}{2}}} \\ \end{gathered}$

Taking the limit of both sides as $\Delta x \to 0$, we have
$\begin{gathered} \Rightarrow \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = – \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin \left( {\frac{{2x + \Delta x}}{2}} \right)\sin \left( {\frac{{\Delta x}}{2}} \right)}}{{\frac{{\Delta x}}{2}}} \\ \Rightarrow \frac{{dy}}{{dx}} = \mathop { – \lim }\limits_{\Delta x \to 0} \sin \left( {\frac{{2x + \Delta x}}{2}} \right)\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin \left( {\frac{{\Delta x}}{2}} \right)}}{{\frac{{\Delta x}}{2}}} \\ \end{gathered}$

Considering $\frac{{\Delta x}}{2} = u$, as $\Delta x \to 0$ then $u \to 0$, we get
$\Rightarrow \frac{{dy}}{{dx}} = – \mathop {\lim }\limits_{\Delta x \to 0} \sin \left( {\frac{{2x + \Delta x}}{2}} \right)\mathop {\lim }\limits_{u \to 0} \frac{{\sin u}}{u}$

Using the relation from the limit $\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1$, we have
$\begin{gathered} \Rightarrow \frac{{dy}}{{dx}} = – \sin \left( x \right)\left( 1 \right) \\ \Rightarrow \frac{{dy}}{{dx}} = – \sin x \\ \end{gathered}$

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