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}