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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