Derivative of Cosine

We shall prove formula for derivative of cosine function using by definition or first principle method.
Let us suppose that the function 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 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 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}


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} \sin \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}} = - \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}


Differentiation 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}

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