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प्रश्न
Differentiate the following w.r.t. x :
`sin^-1(4^(x + 1/2)/(1 + 2^(4x)))`
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उत्तर
Let `y = sin^-1(4^(x + 1/2)/(1 + 2^(4x)))`
`y = sin^-1[(4^x.4^(1/2))/(1 + (2^2)^(2x))]`
`y = sin^-1((2.4^x)/(1 + 4^(2x)))`
Put 4x = tan θ, Then θ = tan–1(4x)
∴ `y = sin^-1((2tanθ)/(1 + tan^2 θ))`
∴ y = sin–1(sin 2θ)
∴ y = 2θ
∴ y = 2tan–1 (4x)
Differentiating w.r.t. x, we get,
`dy/dx = d/dx [2tan^-1 (4^x)]`
`dy/dx = 2 d/dx [tan^-1(4^x)]`
`dy/dx = 2 × 1/(1 + (4^x)^2). d/dx (4^x)`
`dy/dx = (2)/(1 + 4^(2x)) xx 4^xlog4`
`dy/dx = (2.4^xlog4)/(1 + 4^(2x))`
Note: The answer can also be written as :
`dy/dx = (4^(1/2).4^xlog4)/(1 + 4^(2x))`
`dy/dx = (4^(x + 1/2).log4)/(1 + 4^(2x))`
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