Consider the function f(x) = sin^-1(3x-4x³) + cos^-1(4x³-3x)

Let the dreivative of the function be, f'(x) = D₁+D₂

where D₁ = (d/dx)(sin^-1(3x-4x³)) -------------(1)

and D₂ = (d/dx)(cos^-1(4x³-3x)) ------------(2)

Let x = sinθ so that (1) becomes;

D₁ = (d/dx)(sin^-1(3sinθ-4sin³θ))

ie, D₁ = (d/dx)(sin^-1(sin 3θ)

ie, D₁ = (d/dx)(3θ)

ie, D₁ = (d/dx) (3 sin^-1 x)

similarly, putting x = cos φ in (2), we can show that

D₂ = (d/dx) (3 cos^-1 x)

Therefore, f'(x) = D₁+D₂

= (d/dx) (3 sin^-1 x) + (d/dx) (3 cos^-1 x)

= (d/dx)(3(sin^-1 x+ cos^-1 x)

= (d/dx)(3π÷2)

=0

But, We can prove that the aboce given function f(x) is not a constant function either by substituting 3x-4x³ = y or using a scientific calculator!!!

How this is possible? Can the derivative of a non-constant function be zero???

dude, there is no rule that the derivative of a function is not zero it is always zero for a constant function one thing is that you have definitely gone wrong when you have substituted 3theta for arcsin[sin[theta]] by the way may i know the values of your scientific calculator that has shown dif values

Dear Davis Devasia

sin^-1 x+ cos^-1 x =∏/2

you can use this formula if x is same in both sin^-1x and cos^-1x

but here you can use it because x in sin^-1x is sinΘ  

and x in cos^-1x is cosΦ   ,both are different .you can use this formula

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