Question

What is the derivative of `y=arcsin(2x+1)`?

Answer 1

The derivative is `=1/sqrt(-x(x+1))`

Explanation:

`y=arcsin(2x+1)`

So,

`siny=2x+1`

Differentiating wrt `x`

`cosy*dy/dx=2`

`dy/dx=2/cosy`

`sin^2y+cos^2y=1`

`cos^2y=1-sin^2y`

`=1-(2x+1)^2`

`=1-4x^2-4x-1`

`=-4(x^2+x)`

`cosy=sqrt(-4(x^2+x))`

Therefore,

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

`=1/sqrt(-x(x+1))`

Answer 2

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

Explanation:

`color(orange)"Reminder " d/dx(sin^-1x)=1/sqrt(1-x^2)`

`"differentiate using the "color(blue)"chain rule"`

`• d/dx(sin^-1(f(x)))=1/sqrt(1-(f(x))^2)xxf'(x)`

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

`rArrdy/dx=1/sqrt(1-(2x+1)^2)xxd/dx(2x+1)`

`color(white)(rArrdy/dx)=2/sqrt(1-(2x+1)^2)`