calculus inverse derivatives

Question

find f'(x): 

f(x)=tan^-1(x^4)+(tanx^4)^-1+(tan^-1x)^4

Answer 1

f(x) = tan^-1(x⁴) + (tan x⁴)^-1 + (tan^-1 x)⁴

 

The first term is the derivative of inverse tangent, with a chain rule inside.

The second term is a power rule, with a chain rule of regular tangent inside.

The third term is a power rule, with a chain rule of inverse tangent inside.

 

f'(x) = 1 / (1+(x⁴)²) * (4x³) + (-1) (tan x⁴)^-2 (sec² x⁴) (4x³) + 4 (tan-1 x)³ * [1 / (1 + x²)]

f'(x) = 4x³ / (1 + x⁸) - (4x³ sec² x⁴) / (tan² x⁴) + 4 (tan^-1 x)³ / (1 + x²)

f'(x) = 4x³ / (1 + x⁸) - 4x³ csc² x⁴ + 4 (tan^-1 x)³ / (1 + x²)

Answer 2

f'(x)= (4x^3)/(1+x^8) -(4x^3)/(1+x^8)  + 4((tan^-1(x)^3)/(1+x^2)

 

      = 4((tan^-1(x)^3)/(1+x^2)