# Show algebraically that G'(x) = F'(x). Let `G(x)= 2sin^(-1)((sqrt(x))/2)` Let `F(x) = sin^(-1)((x-2)/2)` Please show every single step, without using any shortcuts

*print*Print*list*Cite

### 1 Answer

`G(x)= 2 sin^(-1) (sqrtx/2)`

`F(x)=sin^(-1)((x-2)/2)`

To show that G'(x)=F'(x), determine the derivative of each given function. Use the formula:

`d/(du) sin^(-1)u = 1/sqrt(1-u^2)*u'`

So, the derivative of G(x) is:

`G(x)=2sin^(-1) (sqrtx/2)`

`G'(x) =2*1/sqrt(1-(sqrtx/2)^2)* (sqrtx/2)'`

`G'(x)=2/sqrt(1-x/4)*1/(4sqrtx)=2/(4sqrtxsqrt((4-x)/4))=2/(4sqrtx*1/2sqrt(4-x))`

`G'(x)=1/(sqrtxsqrt(4-x))`

To simplify, use the rule of radicals which is `root(n)(x)*root(n)(y) = root(n)(x*y)` .

`G'(x)=1/sqrt(x(1-x))`

And, the derivative of F'(x) is:

`F(x)=sin^(-1)((x-2)/2)`

`F'(x)=1/sqrt(1-((x-2)/2)^2) *((x-2)/2)'`

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

`F'(x)=1/sqrt(x(4-x))`

Substituting the derivative of the two functions to G'(x) = F(x) yields:

`G'(x)=F'(x)`

`1/sqrt(x(4-x)) = 1/sqrt(x(4-x))` (True)

**Hence this prove that G'(x)=F'(x) .**