Question

D/d{cosec^-1(1+x²/2x)} is equal to​

Answer 1

Step-by-step explanation:

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

$\rm :\longmapsto\:\dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 + {x}^{2} }{2x} \bigg)$

Let assume that

$\rm :\longmapsto\:y = {cosec}^{ - 1} \bigg( \dfrac{1 + {x}^{2} }{2x} \bigg)$

We know,

$\boxed{\tt{ {cosec}^{ - 1}x = {sin}^{ - 1}\bigg( \dfrac{1}{x} \bigg)}}$

So, using this, we get

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

Now, we use Method of Substitution, So we substitute

$\red{\rm :\longmapsto\:x = tanz \: \rm\implies \:z = {tan}^{ - 1}x}$

So, above expression can be rewritten as

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

$\rm :\longmapsto\:y = sin^{ - 1} \bigg( sin2z \bigg)$

$\rm\implies \:y = 2z$

$\bf\implies \:y = 2 {tan}^{ - 1}x$

So,

$\bf\implies \: {cosec}^{ - 1}\bigg( \dfrac{1 + {x}^{2} }{2x} \bigg) = 2 {tan}^{ - 1}x$

Thus,

$\rm :\longmapsto\:\dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 + {x}^{2} }{2x} \bigg)$

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

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

$\rm \: = \: 2 \times \dfrac{1}{1 + {x}^{2} }$

$\rm \: = \: \dfrac{2}{1 + {x}^{2} }$

Hence,

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

Hence, Option (d) is correct.