Q21. Differentiate sinx÷xfrom first principles.

1.	Q21. Differentiate sinx÷xfrom first principles.
Answer» ▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬ Question : [tex]\large\red\bigstar\sf{ Differentiate \: \dfrac{sinx}{x} \: from \: FIRST \: principle }[/tex] Formula for Differentiation : $\\ \large \star \sf \: \frac{d}{dx} sinx = <klux>COSX</klux>$ $\\ \star \large \sf \: \frac{d}{dx} x = 1$ $\\ \star \large \sf \: \frac{d}{dx} \frac{sinx}{x} = \frac{x \frac{d}{dx}sinx - \frac{d}{dx}xsinx }{ {x}^{2} } = \frac{xcosx - sinx}{ {x}^{2} } ....(by \: chain \: rule)$ Formulas for Differentiation from first Principle : $\large\red\bigstar\boxed{\sf{ \dfrac{dy}{dx} = lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}}}$ Formulas of trigonometry: $\\ \large \red \bigstar \sf \: sin(x + y) = 2sin \frac{(x + y)}{2} cos \frac{(x - y)}{2}$ $\\ \red \bigstar \large \sf \: sin( x - y) = 2cos \frac{(x - y)}{2} sin \frac{(x + y)}{2}$ $\\ \red \bigstar \large \sf \: cos(x + y) = 2cos \frac{(x + y)}{2} cos \frac{(x - y)}{2}$ $\\ \red \bigstar \large \sf \: cos(x - y) = - 2sin \frac{(x + y)}{2} sin \frac{(x - y)}{2}$ Solution : ➡ To FIND the derivative of sinx, we use first principle of differentiation. consider the function y=f(x) $\\ \implies\large\sf{ \dfrac{dy}{dx} = lim_{ h \to 0} \: \dfrac{f(x+h)-f(x)}{h}}$ Substituting y=sinx we get, $\\ \implies\large\sf{ \dfrac{dy}{dx} = lim_{ h \to 0} \: \dfrac{sin(x+h)-sinx}{h}}$ Applying the sine difference formula , we have $\\ \implies\large\sf{ \dfrac{dy}{dx} =lim_{h \to 0} \: \dfrac{sin(x+h)-sinx}{h}}$ $\\ \implies\large\sf{ \dfrac{dy}{dx} = lim_{ h \to 0 } \: \dfrac{2cos(x+\dfrac{h}{2}) sin\dfrac{h}{2}}{h}}$ We can PUT $\sf{\dfrac{h}{2}}$ .Then u =0 and h=0 so the limits becomes $\\ \implies\large\sf{ \dfrac{dy}{dx} = lim_{h \to 0 } \: \dfrac{2cos(x+u)sinu}{2u}}$ $\\ \implies\large\sf{ \dfrac{dy}{dx} = \bigg(lim_{h \to 0} \: cos(x+u) \bigg) \times \bigg(lim_{h \to 0} \: \dfrac{sinu}{u} \bigg) }$ $\\ \implies\large\sf{ \dfrac{dy}{dx} =cosx \times 1 }$ $\\ \implies\large\sf{ \dfrac{dy}{dx} =cosx}$ We can conclude that $\\ \implies \large \sf \: \frac{d}{dx} sinx = cosx$ The differentiation of sinx from first principle is cosx.

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Question :

[tex]\large\red\bigstar\sf{ Differentiate \: \dfrac{sinx}{x} \: from \: FIRST \: principle }[/tex]

Formula for Differentiation :

$\\ \large \star \sf \: \frac{d}{dx} sinx = <klux>COSX</klux>$

$\\ \star \large \sf \: \frac{d}{dx} x = 1$

$\\ \star \large \sf \: \frac{d}{dx} \frac{sinx}{x} = \frac{x \frac{d}{dx}sinx - \frac{d}{dx}xsinx }{ {x}^{2} } = \frac{xcosx - sinx}{ {x}^{2} } ....(by \: chain \: rule)$

Formulas for Differentiation from first Principle :

$\large\red\bigstar\boxed{\sf{ \dfrac{dy}{dx} = lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}}}$

Formulas of trigonometry:

$\\ \large \red \bigstar \sf \: sin(x + y) = 2sin \frac{(x + y)}{2} cos \frac{(x - y)}{2}$

$\\ \red \bigstar \large \sf \: sin( x - y) = 2cos \frac{(x - y)}{2} sin \frac{(x + y)}{2}$

$\\ \red \bigstar \large \sf \: cos(x + y) = 2cos \frac{(x + y)}{2} cos \frac{(x - y)}{2}$

$\\ \red \bigstar \large \sf \: cos(x - y) = - 2sin \frac{(x + y)}{2} sin \frac{(x - y)}{2}$

Solution :

➡ To FIND the derivative of sinx, we use first principle of differentiation. consider the function y=f(x)

$\\ \implies\large\sf{ \dfrac{dy}{dx} = lim_{ h \to 0} \: \dfrac{f(x+h)-f(x)}{h}}$

Substituting y=sinx we get,

$\\ \implies\large\sf{ \dfrac{dy}{dx} = lim_{ h \to 0} \: \dfrac{sin(x+h)-sinx}{h}}$

Applying the sine difference formula , we have

$\\ \implies\large\sf{ \dfrac{dy}{dx} =lim_{h \to 0} \: \dfrac{sin(x+h)-sinx}{h}}$

$\\ \implies\large\sf{ \dfrac{dy}{dx} = lim_{ h \to 0 } \: \dfrac{2cos(x+\dfrac{h}{2}) sin\dfrac{h}{2}}{h}}$

We can PUT $\sf{\dfrac{h}{2}}$ .Then u =0 and h=0 so the limits becomes

$\\ \implies\large\sf{ \dfrac{dy}{dx} = lim_{h \to 0 } \: \dfrac{2cos(x+u)sinu}{2u}}$

$\\ \implies\large\sf{ \dfrac{dy}{dx} = \bigg(lim_{h \to 0} \: cos(x+u) \bigg) \times \bigg(lim_{h \to 0} \: \dfrac{sinu}{u} \bigg) }$

$\\ \implies\large\sf{ \dfrac{dy}{dx} =cosx \times 1 }$

$\\ \implies\large\sf{ \dfrac{dy}{dx} =cosx}$

We can conclude that

$\\ \implies \large \sf \: \frac{d}{dx} sinx = cosx$