Question

if `y = sin^(-1)[(6x-4sqrt(1-4x^2))/5]` Find `dy/dx `.

Answer 1

Given that

`y = sin^(-1)[(6x-4sqrt(1-4x^2))/5]`

if y = sin^-1x, then `dy/dx=1/sqrt(1-x^2)`

`y = sin^(-1)[6x-4sqrt(1-4x^2)/5]`

`=> y = sin^(-1)[(6x)/5-(4sqrt(1-4x^2))/5]`

`=> y =sin^(-1)[(2x xx3)/5-(4sqrt(1-(2x)^2))/5]`

`=>y = sin^(-1)[2xx3/5-4/5sqrt(1-(2x)^2)]`

`=>y = sin^(-1)[2xxsqrt(1-(4/5)^2)-4/5sqrt(1-(2x)^2)]`

We know that

`sin^(-1)p-sin^(-1)q=sin^-1(psqrt(1-q^2)-qsqrt(1-p^2)))`

Here, p=2x and  `q=4/5`

Therefore,

`y= sin^(-1)2x-sin^(-1)`

Differentiating the above function with respect to x, we have,

 `dy/dx=1/sqrt(1-(2x)^2)xx2-0`

 `=>dy/dx=2/sqrt(1-4x^2)`