Calculating the Derivative of arcsin(x) and Evaluating at x = 1/2 | A Step-by-Step Guide

ddx(sin−1x)∣∣x=12=

To find the derivative of the function sin⁻¹x with respect to x, we can use the chain rule

To find the derivative of the function sin⁻¹x with respect to x, we can use the chain rule. The chain rule states that if we have a composition of functions, y = f(g(x)), then the derivative of y with respect to x is given by dy/dx = f'(g(x)) * g'(x).

In this case, sin⁻¹x is the composition of two functions. Let y = sin⁻¹x. Then we can rewrite this as y = arcsin(x), where y = g(x) and g(x) = x. The inner function is g(x) = x and the outer function is f(t) = arcsin(t).

First, we need to find the derivative of the inner function, g'(x). The derivative of x with respect to x is 1, so g'(x) = 1.

Next, we need to find the derivative of the outer function, f'(t). The derivative of arcsin(t) is given by f'(t) = 1/√(1 – t²). Since our variable is x, we replace t with x. Therefore, f'(x) = 1/√(1 – x²).

Now, we can apply the chain rule using the derivatives we found:

dy/dx = f'(g(x)) * g'(x)
= (1/√(1 – x²)) * 1
= 1/√(1 – x²)

Finally, to find the value of the derivative at x = 1/2, we substitute x = 1/2 into the derived expression:

dy/dx∣∣x=1/2 = 1/√(1 – (1/2)²)
= 1/√(1 – 1/4)
= 1/√(3/4)
= 1/(√3/2)
= 2/√3

Therefore, d/dx(sin⁻¹x)∣∣x=1/2 = 2/√3.

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