Question

Differentiate the following w.r.t. x :

`tan^(−1)[(2^(x + 2))/(1 − 3(4^x))]`

Answer 1

Let `y = tan^(−1)[(2^(x + 2))/(1 − 3(4^x))]`

`y = tan^(−1)[(2^2 . 2^x)/(1 − 3(4^x))]`

`y = tan^(−1)[(4.2^x)/(1 − 3(2^x)^2)]`

`y = tan^(−1)[(3 × 2^x + 1 × 2^x)/(1 − 3.2^x × 1.2^x)]    ...[tan^(−1) x + tan^(−1) y = tan^(-1) ((x + y)/(1 - xy))]`

y = tan^–1(3.2^x) + tan^–1(2^x)

Differentiating w.r.t. x, we get

`dy/dx = d/dx [tan^-1 (3.2^x) + tan^-1(2^x)]`

`dy/dx = d/dx [tan^-1(3.2^x)] + d/dx [tan^-1 (2^x)]`

`dy/dx = (1)/(1 + (3.2^x)^2). d/dx (3.2^x) + (1)/(1 + (2^x)^2). d/dx (2^x)`

`dy/dx = (1)/(1 + 9(2^(2x))) × 3 × 2^xlog2 + (1)/(1 + 2^(2x)) × 2^xlog2`

`dy/dx = 2^xlog2[(3)/(1 + 9(2^(2x))) + (1)/(1 + 2^(2x))]`