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???
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???