Consider the function f(x) = sin -1 (3x-4x 3 ) + cos -1 (4x 3 -3x) Let the dreivative of the function be, f'(x) = D 1 +D 2 where D 1 = (d/dx)(sin -1 (3x-4x 3 )) -------------(1) and D 2 = (d/dx)(cos -1 (4x 3 -3x)) ------------(2) Let x = sinθ so that (1) becomes; D 1 = (d/dx)(sin -1 (3sinθ-4sin 3 θ)) ie, D 1 = (d/dx)(sin -1 (sin 3θ) ie, D 1 = (d/dx)(3θ) ie, D 1 = (d/dx) (3 sin -1 x) similarly, putting x = cos φ in (2), we can show that D 2 = (d/dx) (3 cos -1 x) Therefore, f'(x) = D 1 +D 2 = (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-4x 3 = y or using a scientific calculator!!! How this is possible? Can the derivative of a non-constant function be zero???

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???