## ddx(sin−1x)∣∣x=12=

### To find the derivative of arcsin(x) and evaluate it at x = 1/2, let’s start by recalling the definition and properties of the arcsine function

To find the derivative of arcsin(x) and evaluate it at x = 1/2, let’s start by recalling the definition and properties of the arcsine function.

The arcsin(x) function is the inverse of the sine function. It takes an input value x and returns the angle (in radians) whose sine is x. The domain of arcsin(x) is [-1, 1], and the range is [-π/2, π/2].

Now, let’s find the derivative of arcsin(x) using the chain rule:

1. Begin with the function y = arcsin(x).

2. Take the sine of both sides: sin(y) = x.

3. Differentiate both sides with respect to x:

d/dx(sin(y)) = d/dx(x),

cos(y) * dy/dx = 1.

4. Solve for dy/dx:

dy/dx = 1 / cos(y).

5. Substitute back the original equation sin(y) = x:

dy/dx = 1 / cos(arcsin(x)).

Now, we can evaluate this derivative at x = 1/2:

dy/dx ∣∣ x=1/2 = 1 / cos(arcsin(1/2)).

Since arcsin(1/2) = π/6 (or 30 degrees), we have:

dy/dx ∣∣ x=1/2 = 1 / cos(π/6).

The value of cos(π/6) is √3/2, so we can substitute it in:

dy/dx ∣∣ x=1/2 = 1 / (√3/2).

To simplify further, we multiply the fraction by its conjugate:

dy/dx ∣∣ x=1/2 = 2 / √3.

To rationalize the denominator, multiply the fraction by √3/√3:

dy/dx ∣∣ x=1/2 = 2√3 / 3.

Therefore, the derivative of arcsin(x) evaluated at x = 1/2 is 2√3 / 3.

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