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The chain rule is a fundamental theorem in calculus used to differentiate composite functions. Imagine you have a function nested within another. The chain rule helps us find the derivative by chaining through these layers.

If you have a function like \( y = f(g(x)) \), the chain rule states that the derivative \( \frac{dy}{dx} \) is the product of the derivative of the outer function \( f \) with respect to \( g \), and the derivative of \( g \) with respect to \( x \). This is often written as:

• \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \)

For our exercise, \( y = \sin^{-1}(u) \) is a perfect example of where the chain rule is useful. Here, \( u \) is a function of \( x \), specifically \( \frac{x}{\sqrt{1+x^2}} \). To find the derivative of \( y \), you first find the derivative of the arcsine function with respect to \( u \) and multiply it by the derivative of \( u \) with respect to \( x \). This layered differentiation process is exactly what the chain rule is all about.

Inverse trigonometric functions, like sine inverse \( \sin^{-1}(x) \), are functions that reverse the effect of trigonometric functions. For example, while \( \sin(\theta) \) tells us the sine of an angle \( \theta \), \( \sin^{-1}(x) \) tells us the angle whose sine is \( x \).

When differentiating inverse trigonometric functions, each has its derivative formula. Knowing these formulas is critical for calculus. For \( \sin^{-1}(x) \), the derivative with respect to \( x \) is:

• \( \frac{1}{\sqrt{1-x^2}} \)

In our exercise, you must apply this derivative rule, but since \( \sin^{-1} \) is applied to a complex expression, the chain rule must be used simultaneously. This combination ensures you train not only in inverse derivatives but in combining multiple differentiation techniques.

The quotient rule is another pivotal differentiation technique used when dealing with fractions. It advises how to find the derivative of a quotient of two functions, say \( \frac{v}{w} \).

The formula for the quotient rule is:

• \( \frac{d}{dx}\left(\frac{v}{w}\right) = \frac{w \cdot \frac{dv}{dx} - v \cdot \frac{dw}{dx}}{w^2} \)

In the exercise, \( u = \frac{x}{\sqrt{1+x^2}} \) needs differentiation. You select \( v = x \) and \( w = \sqrt{1+x^2} \), and apply the quotient rule.

This involves different steps: finding the derivatives \( \frac{dv}{dx} \) and \( \frac{dw}{dx} \), plugging them into the rule, and simplifying the result. The rule elegantly handles the complexity of differentiating fractions, which are often not straightforward.

The arcsine function \( \sin^{-1}(x) \) returns the angle whose sine is \( x \). It's one of the several inverse trigonometric functions you'll encounter.

When tasked with differentiating \( \sin^{-1}(x) \), remember this derivative structure is:

• \( \frac{d}{dx}[\sin^{-1}(x)] = \frac{1}{\sqrt{1-x^2}} \)

In our problem, since \( \sin^{-1} \) is applied to a compound function \( \frac{x}{\sqrt{1+x^2}} \), it requires application of both the derivative of arcsine and the chain rule. However, during the solution, you eventually find this expression simplifies to \( \arctan(x) \), whose derivative conveniently is \( 1 \). This shows how functions can transform through algebra into simpler, well-known derivatives, reaffirming the power and elegance of calculus techniques.