Question

Discuss the differentiability of the function

$f(x) = {sin}^{ - 1} \frac{2x}{1 + {x}^{2} }$
​

Answer 1

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Answer 2

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

Given that

$\rm :\longmapsto\:f(x) = {sin}^{ - 1}\bigg[\dfrac{2x}{1 + {x}^{2} } \bigg]$

On differentiating both sides, w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}f(x) =\dfrac{d}{dx} {sin}^{ - 1}\bigg[\dfrac{2x}{1 + {x}^{2} } \bigg]$

We know,

$\boxed{ \rm \:\dfrac{d}{dx} {sin}^{ - 1}x = \frac{1}{ \sqrt{1 - {x}^{2} } }}$

So, using this, we get

$\rm :\longmapsto\:f'(x) = \dfrac{1}{ \sqrt{1 - {\bigg[\dfrac{2x}{1 + {x}^{2} } \bigg]}^{2} } }\dfrac{d}{dx}\bigg[\dfrac{2x}{1 + {x}^{2} } \bigg]$

We know,

$\boxed{ \rm \:\dfrac{d}{dx} \frac{u}{v} = \frac{v\dfrac{d}{dx}u - u\dfrac{d}{dx}v}{ {v}^{2} } }$

So, using this

$\rm \: = \dfrac{1}{ \sqrt{1 - \dfrac{ {4x}^{2} }{ {(1 + {x}^{2}) }^{2} } } } \bigg[\dfrac{(1 + {x}^{2})\dfrac{d}{dx}2x - 2x\dfrac{d}{dx}(1 + {x}^{2})}{ {(1 + {x}^{2} )}^{2} } \bigg]$

We know,

$\boxed{ \rm \:\dfrac{d}{dx}k = 0}$

and

$\boxed{ \rm \:\dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}$

So, using this,

$\rm \: = \dfrac{1 + {x}^{2} }{ \sqrt{ {(1 + {x}^{2}) }^{2} - {4x}^{2} } }\bigg[\dfrac{(1 + {x}^{2})2 - 2x(2x) }{ {(1 + {x}^{2} )}^{2} } \bigg]$

$\red{\bigg \{ \because \sqrt{ {x}^{2} } = |x| \: \: \: and \: |1 + {x}^{2} | = 1 + {x}^{2} \bigg \}}$

$\rm \: = \dfrac{1 }{ \sqrt{ {(1 - {x}^{2}) }^{2}} }\bigg[\dfrac{2 + 2{x}^{2} - {4x}^{2} }{ {(1 + {x}^{2} )}^{} } \bigg]$

$\rm \: = \: \dfrac{1}{ |1 - {x}^{2} | }\bigg[\dfrac{2 - {2x}^{2} }{1 + {x}^{2} } \bigg]$

$\red{\bigg \{ \because \: \sqrt{ {x}^{2} } = |x| \bigg \}}$

$\rm \: = \: \dfrac{1}{ |(1 - x)(1 + x) | }\bigg[\dfrac{2(1 - {x}^{2} )}{1 + {x}^{2} } \bigg]$

Now,

$\red{\rm :\longmapsto\:f'(x) \: exist \: iff \: (1 - x)(1 + x) \: \ne \: 0}$

$\rm :\implies\:f'(x) \: exist \: \forall \: x \: \in \: R \: - \: \{ - 1, \: 1 \}$

$\rm :\implies\:f(x) \: is \: differentiable \: \forall \: x \: \in \: R \: - \: \{ - 1, \: 1 \}$

Additional Information :-

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