Question

Differentiate using first principal

f(x) = cos (2x - 5)

​

Answer 1

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

Given function is

$\rm :\longmapsto\:f(x) = cos(2x - 5)$

So,

$\rm :\longmapsto\:f(x + h) = cos[2(x + h) - 5] = cos(2x + 2h - 5)$

By Definition of First Principal, we have

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

On substituting the values, we get

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{h \to 0} \frac{cos(2x + 2h - 5) - cos(2x - 5)}{h}$

We know,

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

So, on using this identity, we get

$\rm =\displaystyle\lim_{h \to 0} \frac{ - 2sin\bigg[\dfrac{2x + 2h - 5 + 2x - 5}{2} \bigg]sin\bigg[\dfrac{2x + 2h - 5 - 2x + 5}{2} \bigg]}{h}$

$\rm =\displaystyle\lim_{h \to 0} \frac{ - 2sin\bigg[\dfrac{4x + 2h - 10}{2} \bigg]sin\bigg[\dfrac{2h }{2} \bigg]}{h}$

$\rm =\displaystyle\lim_{h \to 0} \frac{ - 2sin\bigg[\dfrac{4x + 2h - 10}{2} \bigg]sin\bigg[h \bigg]}{h}$

$\rm \:  =  \: - 2 \times \displaystyle\lim_{h \to 0}sin\bigg[\dfrac{4x + 2h - 10}{2} \bigg] \times \displaystyle\lim_{h \to 0} \frac{sinh}{h}$

We know,

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

So, using this identity, we get

$\rm \:  =  \: - 2 \times sin\bigg[\dfrac{4x - 10}{2} \bigg] \times 1$

$\rm \:  =  \: - 2sin(2x - 5)$

Hence,

$\red{\rm \implies\:\boxed{ \tt{ \: \dfrac{d}{dx}cos(2x - 5) = - 2sin(2x - 5) \: }}}$

More to know :-

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