Derivative of Sine

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

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

First we take the increment or small change in the function.

\begin{gathered} y + \Delta y = \sin \left( {x + \Delta x} \right) \\ \Delta y = \sin \left( {x + \Delta x} \right) - y \\ \end{gathered}

Putting the value of function y = \sin x in the above equation, we get

\Delta y = \sin \left( {x + \Delta x} \right) - \sin x\,\,\,\,{\text{ - - - }}\left( {\text{i}} \right)

Using formula from trigonometry, we have

\sin A - \sin B = 2\cos \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\cos \left( {\frac{{x + \Delta x + x}}{2}} \right)\sin \left( {\frac{{x + \Delta x - x}}{2}} \right) \\ \Delta y = 2\cos \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\cos \left( {\frac{{2x + \Delta x}}{2}} \right)\sin \left( {\frac{{\Delta x}}{2}} \right)}}{{\Delta x}} \\ \Rightarrow \frac{{\Delta y}}{{\Delta x}} = \frac{{\cos \left( {\frac{{2x + \Delta x}}{2}} \right)\sin \left( {\frac{{\Delta x}}{2}} \right)}}{{\frac{{\Delta x}}{2}}} \\ \end{gathered}

Taking 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{{\cos \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} \cos \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}

Consider \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} \cos \left( {\frac{{2x + \Delta x}}{2}} \right)\mathop {\lim }\limits_{u \to 0} \frac{{\sin u}}{u}

Using the relation from limit \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1, we have

\begin{gathered}\Rightarrow \frac{{dy}}{{dx}} = \cos \left( x \right)\left( 1 \right) \\ \Rightarrow \frac{{dy}}{{dx}} = \cos x \\ \end{gathered}

Example: Find the derivative of

y = f\left( x \right) = \sin 2{x^2}

We have the given function as

y = \sin 2{x^2}

Differentiation with respect to variable x, we get

\frac{{dy}}{{dx}} = \frac{d}{{dx}}\sin 2{x^2}

Using the rule, \frac{d}{{dx}}\sin x = \cos x, we get

\begin{gathered}\frac{{dy}}{{dx}} = \cos 2{x^2}\frac{d}{{dx}}\left( {2{x^2}} \right) \\ \frac{{dy}}{{dx}} = 4x\cos 2{x^2} \\ \end{gathered}