Question

Differentiate the following w.r.t. x :

`sin^-1(4^(x + 1/2)/(1 + 2^(4x)))`

Answer 1

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 4^x = tan θ, Then θ = tan^–1(4^x)

∴ `y = sin^-1((2tanθ)/(1 + tan^2 θ))`

∴ y = sin^–1(sin 2θ)

∴ y = 2θ

∴ y = 2tan^–1 (4^x)

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))`