# Show that arcsin((1-x^2)^1/2)+arcosx=pie? show with help of derivative

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You need to remember that if a function is a constant, then its derivative is zero, hence, you need to differentiate the left side with respect to x such that:

`(arcsin(sqrt(1-x^2))+arccosx)' = (pi)'`

`(1/(sqrt(1- (sqrt(1-x^2))^2)))*((sqrt(1-x^2)))'*(1-x^2)' - 1/(sqrt(1 - x^2)) = 0`

`1/(sqrt(1 - 1 + x^2))*(-2x)/(2sqrt(1-x^2)) - 1/(sqrt(1 - x^2)) = 0`

`1/(sqrt(x^2))*(-x/(sqrt(1-x^2))) - 1/(sqrt(1 - x^2)) = 0`

`1/|x|*(-x/(sqrt(1-x^2))) - 1/(sqrt(1 - x^2)) = 0`

`1/(sqrt(1 - x^2)) - 1/(sqrt(1 - x^2)) = 0`

`0 = 0`

Hence, evaluating the derivative of the left side yields 0, hence, the function `arcsin(sqrt(1-x^2))+arccosx` is a constant.

Now, you should prove that this constant is equal to pi, hence, you may give values to x such that:

`x = 0 => arcsin(sqrt(1-0^2))+arccos0 = arcsin 1 + arccos0 `

`arcsin(sqrt(1-0^2))+arccos0 = pi/2 + pi/2 = 2pi/2 = pi`

**Hence, differentiating the left side with respect to x yields 0 and giving the value 0 to x yields `pi` , thus, the expression `arcsin(sqrt(1-x^2))+arccosx = pi` is valid.**