Question

If y = log tan x to base cot x . log cot x to base tan x + tan^-1 ((4x)/(4 + x^2)) then dy/dx=

Answer 1

Given :   y = log tan x to base cot x . log cot x to base tan x + tan⁻¹ ((4x)/(4 -x²))

To Find  :  dy/dx

Solution:

log tan x to base cot x  = log  (cot x)⁻¹  base cot x  = - 1

log cot x to base tan x  = log (tan x)⁻¹  to base tan x  = - 1

(-1)(-1) = 1

=> y =  1  + tan⁻¹ ((4x)/(4 - x^2))

Let say x = 2tanθ   => θ = tan⁻¹(x/2)

(4x)/(4 + x²)  =  4 * 2 tanθ / ( 4 - 4tan²θ) = 2tanθ/( 1  - tan²2θ)  = tan2θ

tan⁻¹ ( tan2θ)  = 2θ

=  2  tan⁻¹(x/2)

y =  1  +  2  tan⁻¹(x/2)

dy/dx = 0  + 2 {  1/( 1 + (x/2)²) } (1/2)

=> dy/dx =   4/(4 + x²)

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