Question

derivative of ln(1-cos x)​

Answer 1

Answer:

d  ln(1-cos x)  / dx

= d  ln(1-cos x) / d(1-cos x)    * d(1-cos x) /dx

= 1 /(1-cos x) * { 0 - ( - sin x ) }

= sin x  / (1-cos x)  

Answer 2

To differentiate :-

• In ( 1 - cos x )

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Solution :-

We know that,

• $\rm \: \dfrac{d}{dx} \bigg( constant\bigg) = 0$

• $\rm \: \dfrac{d}{dx} \bigg( cos \: x\bigg) = - sin \: x$

• $\rm\: \dfrac{d}{dx}\bigg(In\:x \bigg)=\dfrac{1}{x}$

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Apply chain rule [ f(g(x)) ]=f'(g(x)) g'(x)

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$\rm \implies\dfrac{d}{dx} \bigg( ln \: (1 - \cos \: x) \bigg)= \dfrac{1}{1 - \cos \: x } \times \dfrac{d}{dx} \bigg(1 - \cos \: x\bigg)$

$\rm \implies\dfrac{d}{dx} \bigg(ln \: (1 - \cos \: x)\bigg) = \dfrac{1}{1 - \cos \: x } \times 0 - ( - \sin \: x)$

$\rm \implies\dfrac{d}{dx} \bigg( ln \: (1 - \cos \: x) \bigg)= \dfrac{1}{1 - \cos \: x } \times( 0 + \sin \: x)$

$\rm \implies\dfrac{d}{dx} \bigg( ln \: (1 - \cos \: x) \bigg)= \dfrac{1}{1 - \cos \: x } \times( \sin \: x)$

$\rm \implies \: \dfrac{d}{dx} \bigg( ln \: (1 - \cos \: x) \bigg) = \dfrac{ \sin \: x }{1 - \cos \: x }$

Hence the derivative of In (1 - cos x ) = sin x / 1 - cos x.

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More formulas :-

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$\bullet \rm \: \dfrac{d}{dx} \bigg( sin \: x\bigg) = cos \: x$

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$\bullet \rm \: \dfrac{d}{dx} \bigg( tan \: x\bigg) = sec ^{2} \: x$

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$\bullet \rm \: \dfrac{d}{dx} \bigg( sec \: x\bigg) = sec \: x.tan \: x$

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$\bullet \rm \: \dfrac{d}{dx} \bigg( {e}^{x} \bigg) = {e}^{x}$

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$\bullet \rm \: \dfrac{d}{dx} \bigg( \dfrac{1}{x} \bigg) = - \dfrac{1}{ {x}^{2} }$ ‎

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