Problem 16 - Technical Mathematics with Calculus

Question

Find the derivative. $$y=\sin ^{-1} \frac{x}{\sqrt{1+x^{2}}}$$

Step-by-Step Solution

Answer

\frac{dy}{dx} = \sqrt{1+x^2}

Step 1: Apply the inverse trigonometric derivative formula

For an inverse trigonometric function of the form \( y = \sin^{-1}(u) \), the derivative is given by \( \frac{dy}{dx} = \frac{1}{\sqrt{1-u^2}} \frac{du}{dx} \). In our case, \( u = \frac{x}{\sqrt{1+x^2}} \), we need to find \( \frac{du}{dx} \) as well.

Step 2: Differentiate the inner function \( u \)

To differentiate \( u = \frac{x}{\sqrt{1+x^2}} \), apply the quotient rule, \( (\frac{f}{g})' = \frac{f'g - fg'}{g^2} \). Let \( f = x \) and \( g = \sqrt{1+x^2} \), then find \( f' \) and \( g' \) and substitute them into the quotient rule.

Step 3: Compute derivative of \( f = x \) and \( g = \sqrt{1+x^2} \)

\( f' = \frac{d}{dx}x = 1 \) and \( g' = \frac{d}{dx}\sqrt{1+x^2} \). To find \( g' \), use the chain rule: \( g' = \frac{1}{2\sqrt{1+x^2}} \cdot 2x = \frac{x}{\sqrt{1+x^2}} \).

Step 4: Apply the quotient rule to find \( \frac{du}{dx} \)

Using the derivatives of \( f \) and \( g \) from the previous step, apply the quotient rule: \( \frac{du}{dx} = \frac{f'g - fg'}{g^2} = \frac{1 \cdot \sqrt{1+x^2} - x \cdot \frac{x}{\sqrt{1+x^2}}}{1+x^2} \). Simplify this expression to find \( \frac{du}{dx} \).

Step 5: Simplify \( \frac{du}{dx} \)

Simplify the numerator by combining like terms and then simplify the fraction: \( \frac{du}{dx} = \frac{\sqrt{1+x^2} - \frac{x^2}{\sqrt{1+x^2}}}{1+x^2} = \frac{\sqrt{1+x^2} - (x^2) (\frac{1}{\sqrt{1+x^2}})}{1+x^2} = \frac{1}{\sqrt{1+x^2}} \).

Step 6: Combine \( \frac{du}{dx} \) with the outer derivative

Now, plug \( \frac{du}{dx} \) back into the derivative of the outer function to get the final derivative: \( \frac{dy}{dx} = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} \). After substituting \( u \) and \( \frac{du}{dx} \), we find \( \frac{dy}{dx} = \frac{1}{\sqrt{1-\left( \frac{x}{\sqrt{1+x^2}} \right)^2}} \cdot \frac{1}{\sqrt{1+x^2}} \).

Step 7: Simplify the final derivative

Simplify the expression for the final derivative by dealing with the square inside the square root and combining the square roots: \( \frac{dy}{dx} = \frac{1}{\sqrt{1-\frac{x^2}{1+x^2}}} \cdot \frac{1}{\sqrt{1+x^2}} = \frac{\sqrt{1+x^2}}{\sqrt{1+x^2 -x^2}} = \frac{\sqrt{1+x^2}}{\sqrt{1}} = \sqrt{1+x^2} \).

Key Concepts

Chain Rule in Differentiation

The chain rule is a fundamental technique in calculus used to find the derivative of composite functions. When a function can be expressed as one function inside another, the chain rule allows you to differentiate this nested function efficiently.

For instance, if you have a function in the form of \( h(x) = f(g(x)) \), then the chain rule states that the derivative of \( h(x) \) with respect to \( x \) is \( h'(x) = f'(g(x)) \cdot g'(x) \). Essentially, you're taking the derivative of the outer function evaluated at the inner function and then multiplying it by the derivative of the inner function.

In our example with the inverse trigonometric derivative, the chain rule was used to find \( g' \) – the derivative of the inner function \( g = \sqrt{1+x^2} \). The chain rule allowed the simplification of the differentiation process by treating each function individually before combining them to get the final derivative.

Quotient Rule for Derivatives

The quotient rule is a derivative rule that allows us to differentiate expressions where one function is divided by another. It is particularly useful when dealing with ratios of functions, as seen in trigonometric identities and complex fractions.

Mathematically, if we have a function \( u(x) = \frac{f(x)}{g(x)} \), where both \( f \) and \( g \) are differentiable and \( g(x) \) is not zero, the quotient rule states that \( u'(x) = \frac{f'(x) \cdot g(x) - f(x) \cdot g'(x)}{(g(x))^2} \).

Applied to our specific example, the inner function \( u = \frac{x}{\sqrt{1+x^2}} \) is differentiated using the quotient rule, resulting in a more complex derivative that needed to be simplified further to arrive at the final expression.

Differentiation of Trigonometric Functions

Differentiating trigonometric functions is another key concept in calculus. Familiarity with the derivatives of basic trigonometric functions is essential. For example, the derivative of \( \sin(x) \) is \( \cos(x) \) and the derivative of \( \cos(x) \) is \( -\sin(x) \).

However, when we deal with inverse trigonometric functions such as \( \sin^{-1}(x) \) or \( \arcsin(x) \) the derivatives are not as straightforward. The derivative of \( \sin^{-1}(x) \) is \( \frac{1}{\sqrt{1-x^2}} \) under the condition that \( x \) lies within the domain of the function. In the provided exercise, the derivative of the inverse trigonometric function was combined with other differentiation techniques to find the overall derivative of the given composite function.

Simplifying Complex Derivatives

Simplification is an important step in calculus to make a derivative more understandable or to prepare it for further calculations. After applying the chain rule, quotient rule, and the differentiation of trigonometric functions, we often end up with complex expressions that require simplification.

In our exercise, after applying the appropriate rules of differentiation, simplifying complex derivatives helped us arrive at an elegant and simple answer. The final step often involves combining like terms, factoring, and cancelling out terms where possible to achieve the simplest form of a derivative.

Proper simplification not only makes the result more concise but also sheds light on the behavior of the function being differentiated, making it easier to understand and apply in various mathematical and real-world contexts.