Question

Easy Question This time :)

Find the Derivative of Cos x from ab-initio method ! ​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:f(x) = cosx$

So,

$\rm :\longmapsto\:f(x + h) = cos(x + h)$

By definition of First Principal, we have

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{h \to 0}\rm \frac{f(x + h) - f(x)}{h}$

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{h \to 0}\rm \frac{cos(x + h) - cosx}{h}$

We know,

$\boxed{\tt{ cosx - cosy = - 2sin\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg]}}$

So, using this identity, we get

$\rm \:  =  \: \displaystyle\lim_{h \to 0}\rm \frac{ - 2sin\bigg[\dfrac{x + h + x}{2} \bigg]sin\bigg[\dfrac{x + h - x}{2} \bigg]}{h}$

$\rm \:  =  \: \displaystyle\lim_{h \to 0}\rm \frac{ - 2sin\bigg[\dfrac{2x + h}{2} \bigg]sin\bigg[\dfrac{h}{2} \bigg]}{h}$

$\rm \:  =  \: - 2\displaystyle\lim_{h \to 0}\rm sin\bigg[\dfrac{2x + h}{2} \bigg] \times \displaystyle\lim_{h \to 0}\rm \frac{sin \dfrac{h}{2} }{h}$

$\rm \:  =  \: -2sinx \times \displaystyle\lim_{h \to 0}\rm \frac{sin \dfrac{h}{2} }{ \dfrac{h}{2} \times 2}$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ \: \: \displaystyle\lim_{x \to 0}\rm \frac{sinx}{x} \: \: = \: \: 1 \: \: }}}$

So, using this, we get

$\rm \:  =  \: - 2sinx \times \dfrac{1}{2}$

$\rm \:  =  \: - \: sinx$

Hence,

$\\ \purple{\rm\implies \:\boxed{\tt{ \dfrac{d}{dx}cosx \: = \: - \: sinx \: \: }}}$

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More to know

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

Step-by-step explanation:

Solution−

Given function is

$\rm :\longmapsto\:f(x) = cosx:⟼f(x)=cosx$

So,

$\rm :\longmapsto\:f(x + h) = cos(x + h):⟼f(x+h)=cos(x+h)$

By definition of First Principal, we have

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{h \to 0}\rm \frac{f(x + h) - f(x)}{h}:⟼f$

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{h \to 0}\rm \frac{cos(x + h) - cosx}{h}:⟼$

cos(x+h)−cos