Introduction

Understanding the derivative of the inverse sine function is crucial for advanced calculus and real-world applications. The inverse sine function, often denoted as arcsin(x) or sin-1(x), has a range that is limited to ([- frac{pi}{2}, frac{pi}{2}]). Therefore, its derivative plays a vital role in various complex mathematical and engineering problems.

The Derivative of arcsin(x)

The process to derive arcsin(x) involves a step-by-step approach using implicit differentiation and integration by parts. Let's explore this in detail:

Given y arcsin(x), we can write:

siny sin(arcsin(x))

Knowing that siny x, we differentiate both sides implicitly:

[frac{d}{dx}[siny] frac{d}{dx}[x]]

[cos(y) frac{dy}{dx} 1]

[frac{dy}{dx} frac{1}{cos(y)}]

Using the Pythagorean identity:

[sin^2(y) cos^2(y) 1]

[cos^2(y) 1 - sin^2(y)]

[cos(y) sqrt{1 - sin^2(y)}]

Substituting , we get:

[cos(y) sqrt{1 - x^2}]

Therefore, the derivative arcsin(x) is:

[frac{dy}{dx} frac{1}{sqrt{1 - x^2}}]

Integration of arcsin(x)

Now, let's explore the integration of arcsin(x). Consider the integral:

[int arcsin(x), dx]

Using integration by parts:

[I int arcsin(x), dx xarcsin(x) - int x cdot frac{1}{sqrt{1 - x^2}}, dx]

Let u arcsin(x) and dv dx. Then du frac{1}{sqrt{1 - x^2}}dx and v x. Thus:

[I xarcsin(x) - int frac{x}{sqrt{1 - x^2}}, dx]

For the integral int frac{x}{sqrt{1 - x^2}}, dx, let u 1 - x^2. Then du -2x, dx, and the integral becomes:

[int frac{x}{sqrt{1 - x^2}}, dx -frac{1}{2} int u^{-frac{1}{2}}, du]

[ -frac{1}{2} cdot 2sqrt{u} -sqrt{1 - x^2}]

Hence, the whole integral is:

[int arcsin(x), dx xarcsin(x) - (-sqrt{1 - x^2}) C]

[ xarcsin(x) sqrt{1 - x^2} C]

This result aligns with the earlier manual calculation and the generalized formula for integrating functions of the form int f^{-1}(x)dx.

Conclusion

The derivative of the inverse sine function, arcsin(x), is a fundamental concept in calculus, and its integral is crucial for applications in various fields. Understanding these derivatives and integrals helps in solving complex mathematical problems and real-world applications involving trigonometric functions.

For further inquiries or deeper understanding, consult advanced calculus textbooks or online resources dedicated to differential and integral calculus.