Exploring the Derivative of the Inverse Sine Function and its Application in Calculus

sin^-1(x)’

To find the derivative of the inverse sine function denoted as sin^-1(x) or arcsin(x), we can use differentiation techniques

To find the derivative of the inverse sine function denoted as sin^-1(x) or arcsin(x), we can use differentiation techniques. Let’s denote the inverse sine function as y = sin^-1(x).

To differentiate y with respect to x, we need to use the chain rule. The chain rule states that if y = f(g(x)), then dy/dx = f'(g(x)) * g'(x). In this case, f(x) = sin^-1(x) and g(x) = x.

Applying the chain rule, we have:
dy/dx = (d/dx) [sin^-1(x)]
= [d/dx) [sin^-1(x) * 1]
= cos(sin^-1(x)) * (d/dx) [sin^-1(x)]

Now, to further simplify the expression, we need to use trigonometric identities. The identity we will use is sin^2θ + cos^2θ = 1.

Since y = sin^-1(x), we know that sin(y) = x.
Rearranging this equation, we have sin^2(y) = x^2.
Using the identity, we can replace sin^2(y) with 1 – cos^2(y).
So, 1 – cos^2(y) = x^2.

Now, we can differentiate both sides with respect to x:
(d/dx) [1 – cos^2(y)] = (d/dx) [x^2]
0 – 2cos(y) * (d/dx) [cos(y)] = 2x

Since we are looking for dy/dx, we can substitute cos(y) = √(1 – sin^2(y)) using the Pythagorean identity:
-2cos(y) * (d/dx) [cos(y)] = 2x
-2√(1 – sin^2(y)) * (d/dx) [√(1 – sin^2(y))] = 2x
-2√(1 – x^2) * (d/dx) [√(1 – x^2)] = 2x

Now, solving for (d/dx) [√(1 – x^2)]:
(d/dx) [√(1 – x^2)] = -x / √(1 – x^2)

Substituting this result back into the equation, we have:
-2√(1 – x^2) * (-x / √(1 – x^2)) = 2x
2x = 2x

Hence, we see that dy/dx = 1 for the inverse sine function sin^-1(x).

Therefore, the derivative of sin^-1(x) is equal to 1.

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