Question

How do you differentiate `y=cos^-1(1-2x^2)`?

Answer 1

`dy/dx = 2/sqrt(1-x^2)`

Explanation:

Using the`color(blue)" chain rule "`

`d/dx[f(g(x)] = f'(g(x)) . g'(x)`

and the standard derivative `d/dx(cos^-1 x) =( -1)/sqrt(1 - x^2)`

`rArr dy/dx = (-1)/sqrt(1 - (1-2x^2)^2) d/dx(1-2x^2)`

`= (4x)/sqrt(1-1 +4x^2 - 4x^4) =( 4x)/sqrt(4x^2(1-x^2))`

`= (4x)/(2xsqrt(1-x^2))= 2/sqrt(1-x^2)`

Answer 2

`d y=(4x)/(1-2x^2)^2 sin^-1(1-2x^2) d x`

Explanation:

`y=cos u" " y^'=- u^' * sin u`
`y=cos^-1(1-2x^2)" "y=cos(1/(1-2x^2))`
`d y=-((0*(1-2x^2)+4x*1)/(1-2x^2)^2)sin(1/(1-2x^2))*d x`
`d y=(4x)/(1-2x^2)^2 sin^-1(1-2x^2) d x`