Question

Find derivative using first principal

$f(x) = x \: cosx$
​

Answer 1

Answer:

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Answer 2

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

Given function is

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

So,

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

By Definition of First Principal, we have

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

So, on substituting the values, we get

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

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \frac{xcos(x + h) + hcos(x + h) - xcosx}{h}$

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \frac{xcos(x + h) - xcosx}{h} + \displaystyle\lim_{h \to 0}\rm \frac{hcos(x + h)}{h}$

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

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

We know,

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

So, on using this identity, we get

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

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

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

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

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

We know,

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

So, using this result, we

$\rm \: = \: - \: xsinx \times 1 + cosx$

$\rm \: = \: - \: xsinx + cosx$

Hence,

$\rm\implies \:\boxed{\rm{ \: \: \dfrac{d}{dx}(x \: cosx) = - x \: sinx \: + \: cosx \: }}$

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Additional Information :-

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