Question

10. Find the derivative of cos x from first principle

Answer 1

Step-by-step explanation:—

Here,

f(x) =cos x

we have to find the f'(x)

then

$f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

→ f(x) = cos x

→ f(x+h) = cos (x+h)

Putting the value we get

$f'(x) = \lim_{h\rightarrow 0}\frac{cos (x+h)-cos(x)}{h}</p><p></p><p><strong>Using</strong><strong> the</strong><strong> formula</strong></p><p></p><p>[tex]cos A - cos B = - 2 sin(\frac{A+B}{2}) sin (\frac{A-B}{2})$

$f'(x) = \lim_{h\rightarrow 0}\frac{-2 sin (\frac{x+x+h}{2}). sin(\frac{x+h-x}{2}) }{h}$

$= \begin{gathered}\lim_{h\rightarrow 0}\frac{-2 sin (\frac{2x+h}{2}). sin(\frac{h}</p><p>{2}) }{h}\\\end{gathered}$

On calculating we get,

$\lim_{h\rightarrow 0}{- sin(\frac{2x+h}{2}).1}$

$\lim_{h\rightarrow 0}{- sin(\frac{2x+h}{2})}$

$= \lim_{h\rightarrow 0}{- sin(\frac{2x+0}{2})}$

$= \lim_{h\rightarrow 0}{- sin(\frac{2x}{2})}$

Hence ,

Derivative of cos X is -sin x

Answer 2

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

Given function is

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

So,

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

By using, Definition of First Principle, we have

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

So, on substituting the values, we get

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

We know,

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

So, using this, we get

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

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

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

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

We know,

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

So, using this, we get

$\rm \:  =  \: - \: sinx \times 1$

$\rm \:  =  \: - \: sinx$

Thus,

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

More to know :-

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