Question

y=1/2 cos^-1 (1-4x^3/1+4x^3), find dy/dx​

Answer 1

Answer:

let us try this....

if I try to solve it as function of function, I am going to be in deep trouble I guess. so first I try to make it simpler somehow.

2y = cos^-1 ((1-4x^3)/(1+4x^3))

cos2y = (1-4x^3)/(1+4x°3)

using componendo and dividendo

(1+cos2y)/(1-cos2y) = (1+4x^3+1-4x^3)/(1+4x^3-1+4x^3)

2cos^2y/2sin^2y = 2/8x^3

tan^2y = 4x^3

tany = 2x^(3/2). (I am ignoring -ve for the time being)

y = tan^-1(2x^(3/2)

now,

dy/dx = 1/(1+(2x^(3/2)^2)*2*3/2*x^(3/2-1)

dy/dx = 1/(1+4x^3) * 3*√x

dy/dx = 3√x/(1+4x^3)

now, if we take the -ve sign while calculating tany, it will only add a -ve sign in our result, so we may write our solution as

dy/dx = +_ 3√x/(1+4x^3)