Differentiation - inverse tangent? Help!?

Question

find the derivative y=sqrt(tan^-1 x)

Answer 1

We know that if f is differentiable at x, has and inverse and f'(f^(-1)(x) <>0 (different from 0), then it's inverse f^(-1) is differentiable at x and f^(-1)'(x) = 1/f'(f^(-1)(x).

We know d/dx tan(x) = sec^2(x) for every x and sec^2(x) never vanishes. So, (tan^-1 x)' = 1/(sec^2(tan^(-1)(x)))

But sec(tan^(-1)x) = 1 + x^2, according to that well know trigonometric identity. Therefore, (tan^-1 x)' = 1/(x^2 + 1).

Therefore, if y = sqrt(tan^(-1)(x), then, y' = 1/(2sqrt(tan^(-1)(x)) * (tan^-1 x)' = 1/2sqrt(tan^(-1)(x)*(x^2 + 1))

Answer 2

dy/dx=1/2*SQRT[tan^(-1) x](^-1)*1/(1+x^2)

that's the derivative of a composite function. That is, if by tan^-1 you mean the angle whose tangent is x, the inverse function of the tangent.

If by tan^-1 you mean cos/sin (aka cotan), then:

dy/dx=-1/2*SQRT[tan^(-1) x]^(-1)*(cosec(x))

Answer 3

The direct method is to use the chain rule.

y = sqrt(tan^(-1)(x))

We need to know two derivatives:
1) d/dx sqrt(x) = 1/[2sqrt(x)]
2) d/dx tan^(-1)(x) = 1/[1 + x^2]

dy/dx = ( 1/[2tan^(-1)(x)] ) (1/[1 + x^2])

Answer 4

let t=tan^-1 x

y=sqrt(t)

dy/dt=1/2t^(-1/2) = 1/(2*sqrt(t))

dt/dx = 1/(x²+1)

dy/dx=dy/dt * dt/dx = 1/[2*sqrt(tan^-1 x)*(x²+1)]