Question

Differentiate using first principle.

$\dfrac{1 - cos \: x}{x}$
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Answer 1

ANSWER:

To Do:

• Differentiate (1 - cosx)/x using first principle.

Solution:

We are given that,

$\implies f(x)=\dfrac{1-\cos x}{x}$

We know that, by First Principle,

$\hookrightarrow f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$

So,

$\implies f'(x) =\lim_{h\to0}\dfrac{\left(\dfrac{1-\cos(x+h)}{x+h}\right)-\left(\dfrac{1-\cos x}{x}\right)}{h}$

We know that,

$\hookrightarrow\cos(a+b)=\cos a\cos b-\sin a\sin b$

So,

$\implies f'(x) =\lim_{h\to0}\dfrac{\left(\dfrac{1-\cos x\cosh+\sin x\sin h}{x+h}\right)-\left(\dfrac{1-\cos x}{x}\right)}{h}$

Taking LCM,

$\implies f'(x) =\lim_{h\to0}\dfrac{\left(\dfrac{x(1-\cos x\cosh+\sin x\sin h)-(x+h)(1-\cos x)}{x(x+h)}\right)}{h}$

So,

$\implies f'(x) =\lim_{h\to0}\dfrac{x(1-\cos x\cosh+\sin x\sin h)-(x+h)(1-\cos x)}{xh(x+h)}}$

$\implies f'(x) =\lim_{h\to0}\dfrac{x\!\!\!/\:-x\cos x\cosh+x\sin x\sin h-x\!\!\!/\:-h+x\cos x+h\cos x}{xh(x+h)}$

$\implies f'(x)=\lim_{h\to0}\dfrac{-x\cos x\cosh+x\sin x\sin h-h+x\cos x+h\cos x}{xh(x+h)}$

Taking common,

$\implies f'(x) =\lim_{h\to0}\dfrac{x\cos x(1-\cosh)+x\sin x\sin h-h(1-\cos x)}{xh(x+h)}$

Separating the terms,

$\implies f'(x) =\lim_{h\to0}\dfrac{x\!\!\!/\:\cos x(1-\cosh)}{x \!\!\!/\: h(x+h)}+\dfrac{x \!\!\!/\:\sin x\sin h}{x \!\!\!/\: h(x+h)}-\dfrac{h \!\!\!/\:(1-\cos x)}{xh \!\!\!/\:(x+h)}$

$\implies f'(x) =\lim_{h\to0}\dfrac{\cos x(1-\cosh)}{h(x+h)}+\dfrac{\sin x\sin h}{h(x+h)}-\dfrac{(1-\cos x)}{x(x+h)}$

$\implies f'(x) =\lim_{h\to0}\dfrac{\cos x}{(x+h)}\left(\dfrac{1-\cosh}{h}\right)+\dfrac{\sin x}{(x+h)}\left(\dfrac{\sin h}{h}\right)-\dfrac{(1-\cos x)}{x(x+h)}$

$\implies f'(x) =\lim_{h\to0}\dfrac{\cos x}{(x+h)}\left(\dfrac{1-\cosh}{h}\right)+ \lim_{h\to0}\dfrac{\sin x}{(x+h)}\left(\dfrac{\sin h}{h}\right)-\lim_{h\to0}\dfrac{-(\cos x-1)}{x(x+h)}$$\implies f'(x) =\lim_{h\to0}\dfrac{\cos x}{(x+h)}\left(\dfrac{1-\cosh}{h}\right)+ \lim_{h\to0}\dfrac{\sin x}{(x+h)}\left(\dfrac{\sin h}{h}\right)+\lim_{h\to0}\dfrac{(\cos x-1)}{x(x+h)}$We know that,

$\hookrightarrow\lim_{h\to0}\dfrac{\cos h-1}{h}=0$

$\hookrightarrow\lim_{h\to0}\dfrac{\sin h}{h}=1$

We had,

$\implies f'(x) =\lim_{h\to0}\dfrac{\cos x}{(x+h)}\left(\dfrac{1-\cosh}{h}\right)+ \lim_{h\to0}\dfrac{\sin x}{(x+h)}\left(\dfrac{\sin h}{h}\right)+\lim_{h\to0}\dfrac{(\cos x-1)}{x(x+h)}$Solving the limits,

$\implies f'(x)=\dfrac{\cos x}{(x+0)}(0)+\dfrac{\sin x}{(x+0)}(1)+\dfrac{(\cos x-1)}{x(x+0)}$

$\implies f'(x)=0+\dfrac{\sin x}{x}+\dfrac{(\cos x-1)}{x(x)}$

$\implies f'(x)=\dfrac{\sin x}{x}+\dfrac{\cos x-1}{x^2}$

Taking LCM,

$\implies\bf f'(x)=\dfrac{x\,sin x+cos x-1}{x^2}$