d/d{cosec^-1(1+x²/2x)} is equal to​

Step-by-step explanation:

[tex]\large\underline{\sf{Solution-}}[/tex]

[tex]\rm :\longmapsto\:\dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 + {x}^{2} }{2x} \bigg)[/tex]

Let assume that

[tex]\rm :\longmapsto\:y = {cosec}^{ - 1} \bigg( \dfrac{1 + {x}^{2} }{2x} \bigg)[/tex]

We know,

[tex]\boxed{\tt{ {cosec}^{ - 1}x = {sin}^{ - 1}\bigg( \dfrac{1}{x} \bigg)}}[/tex]

So, using this, we get

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

Now, we use Method of Substitution, So we substitute

[tex] \red{\rm :\longmapsto\:x = tanz \: \rm\implies \:z = {tan}^{ - 1}x}[/tex]

So, above expression can be rewritten as

[tex]\rm :\longmapsto\:y = sin^{ - 1} \bigg( \dfrac{2tanz}{1 + {tan}^{2} z} \bigg)[/tex]

[tex]\rm :\longmapsto\:y = sin^{ - 1} \bigg( sin2z \bigg)[/tex]

[tex]\rm\implies \:y = 2z[/tex]

[tex]\bf\implies \:y = 2 {tan}^{ - 1}x[/tex]

So,

[tex]\bf\implies \: {cosec}^{ - 1}\bigg( \dfrac{1 + {x}^{2} }{2x} \bigg) = 2 {tan}^{ - 1}x[/tex]

Thus,

[tex]\rm :\longmapsto\:\dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 + {x}^{2} }{2x} \bigg)[/tex]

[tex]\rm \:  =  \: \dfrac{d}{dx}(2 {tan}^{ - 1}x)[/tex]

[tex]\rm \:  =  \: 2 \: \dfrac{d}{dx}( {tan}^{ - 1}x)[/tex]

[tex]\rm \:  =  \: 2 \times \dfrac{1}{1 + {x}^{2} } [/tex]

[tex]\rm \:  =  \: \dfrac{2}{1 + {x}^{2} } [/tex]

Hence,

[tex] \purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 + {x}^{2} }{2x} \bigg) = \frac{2}{1 + {x}^{2} }}}}[/tex]

Hence, Option (d) is correct.

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