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LIMITS AND DERIVATIVES — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSLIMITS AND DERIVATIVES1 Mark
Question
Compute the derivative of sin x.
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Answer

Let f(x) = sin x
By using the first principle of derivative, we get
${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{\sin (x + h) - \sin x}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{2\cos \left( {\frac{{x + h + x}}{2}} \right) \cdot \sin \left( {\frac{{x + h - x}}{2}} \right)}}{h}$
$\left[ {\because \sin C - \sin D = 2\cos \left( {\frac{{C + D}}{2}} \right) \times \sin \left( {\frac{{C - D}}{2}} \right)} \right]$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{2\cos \left( {x + \frac{h}{2}} \right) \cdot \sin \frac{h}{2}}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \cos \left( {x + \frac{h}{2}} \right) \cdot \mathop {\lim }\limits_{\frac{h}{2} \to 0} \frac{{\sin \frac{h}{2}}}{{\frac{h}{2}}}\left[ {\begin{array}{*{20}{c}} {{\text{ as }}h \to 0,{\text{ then }}} \\ {\frac{h}{2} \to 0} \end{array}} \right]$
$ = \mathop {\lim }\limits_{h \to 0} \cos \left( {x + \frac{h}{2}} \right) \times 1\quad \left[ {\because \mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \theta }}{\theta } = 1} \right]$
= cos(x + 0) = cos x [putting h = 0]
$\therefore \frac { d } { d x } ( \sin x ) $= cos x

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