Answer 1

`Let `

Putting x=cosθ, we get:

`y =tan^(-1)(sqrt(1-cos^2theta)/costheta)=tan^(-1)(sintheta/costheta)=tan^(-1)(tantheta)=theta`

`y = cos^(−1)x`

On differentiating with respect to x, we get:

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

Now assume that

`z=cos^(-1)(2xsqrt(1-x^2))`

`z=cos^(-1)(2sintheta costheta)=cos^(-1)(sin2theta)=cos^(-1)(cos(pi/2-2theta))=pi/2-2theta=pi/2-2cos^(-1)x`

On differentiating with respect to x, we get:

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

We know that,

`dy/dz=(dy/dx)/(dz/dx)`

So, from equations (1) and (2), we get:

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

Derivative of `tan^(-1)(sqrt(1-x^2)/x)` with respect to `cos^(-1)(2xsqrt(1-x^2))  is -1/2`