Sum

Find `(dy)/(dx) , if y = sin ^(-1) [2^(x +1 )/(1+4^x)]`

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#### Solution

`y = sin^(-1) [(2.2^x)/(1 +(2^x)^2)]`

put 2^{x }= tan θ

`∴ y = sin^(-1) [(2 tan theta ) /(1 + tan^2 theta)]`

= sin^{-1} [ sin 2θ ]

= 2θ

y = 2 tan^{-1} ( 2^{x })

Differentiating wrt x,

`(dy)/(dx) = 2/(1 +(2^x) )xx d/(dx) (2^x)`

`= 2/(1 + (2^x)^2) xx 2^x log 2 = (2 ^ (x+ 1))/(1 + 4^x) log 2 =" sin y log" 2`

Concept: Logarithmic Differentiation

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