Differentiate `sin^(-1)sqrt(1-x^2)` with respect to `cos^(-1)x ,` if (i)`x in (0,1)` (ii) `x in (-1,0)`

Explanation

Given `x in(-1,0)`

However, `x=cos theta`.

`Rightarrow cos theta in(-1,0)` `Rightarrow theta in(frac{pi}{2}, pi)` Hence, `u=sin ^{-1}(sin theta)=pi-theta .` `Rightarrow u=pi-cos ^{-1} x`

On differentiating `u` with respect to `x`, we get

`frac{d u}{d x}=frac{d}{d x}(pi-cos ^{-1} x)`

`Rightarrow frac{d u}{d x}=frac{d}{d x}(pi)-frac{d}{d x}(cos ^{-1} x)`

We know `frac{d}{d {x}}(cos ^{-1} {x})=-frac{1}{sqrt{1-{x}^{2}}}` and derivative of a constant is 0 .

`Rightarrow frac{d u}{d x}=0-(-frac{1}{sqrt{1-x^{2}}})`

`therefore frac{d u}{d x}=frac{1}{sqrt{1-x^{2}}}`

Now, on differentiating `v` with respect to `x`, we get

`frac{d v}{d x}=frac{d}{d x}(cos ^{-1} x)`

`therefore frac{{dv}}{{dx}}=-frac{1}{sqrt{1-{x}^{2}}}`

We have

`frac{d u}{d v}=frac{frac{d u}{d x}}{frac{d x}{d x}}`

`Rightarrow frac{d u}{d v}=frac{sqrt{1-x^{2}}}{-frac{1}{sqrt{1-x^{2}}}}`

`Rightarrow frac{d u}{d v}=frac{1}{sqrt{1-x^{2}}} times(-sqrt{1-x^{2}})`

`therefore frac{d u}{d v}=-1`

Thus,

`frac{d u}{d v}=-1`