Question

Differentiate the following functions: \(\sqrt[3]{\frac{1-x}{(1+x)^{2}}}\)

Short answer

Answer: The derivative of the given function is \(\frac{dy}{dx}=\frac{-1-2x}{3(1+x)^{\frac{10}{3}}}\).

Step-by-step solution

Rewrite the function using properties of exponents

Rewrite the function as a single power: $$y = \sqrt[3]{\frac{1-x}{(1+x)^{2}}} = \left(\frac{1-x}{(1+x)^{2}}\right)^\frac{1}{3}$$ Now we have the function in the format \(y = u^\frac{1}{3}\), where \(u = \frac{1-x}{(1+x)^{2}}\).

Chain Rule

Derivate the function using the Chain Rule: $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$

Calculate \(\frac{dy}{du}\)

Differentiate \(y\) with respect to \(u\), treating \(u\) as the variable $$\frac{dy}{du} = \frac{1}{3}u^{\frac{1}{3}-1} = \frac{1}{3}u^{-\frac{2}{3}}$$

Calculate \(\frac{du}{dx}\)

Differentiate \(u\) with respect to \(x\), using the quotient rule: $$\frac{du}{dx} = \frac{(1+x)^{2} \cdot (-1) - (1-x) \cdot 2(1+x)}{((1+x)^{2})^{2}}$$

Simplify \(\frac{du}{dx}\)

Simplify the expression: $$\frac{du}{dx} = \frac{-1-2x}{(1+x)^{4}}$$

Calculate the derivative \(\frac{dy}{dx}\)

Substitute \(\frac{dy}{du}\) and \(\frac{du}{dx}\) into the Chain Rule equation: $$\frac{dy}{dx} = \frac{1}{3}u^{-\frac{2}{3}} \cdot \frac{-1-2x}{(1+x)^{4}}$$ Replace \(u\) with its original expression: $$\frac{dy}{dx} = \frac{1}{3}\left(\frac{1-x}{(1+x)^{2}}\right)^{-\frac{2}{3}} \cdot \frac{-1-2x}{(1+x)^{4}}$$

Simplify the final expression

Combine and simplify the terms: $$\frac{dy}{dx} = \frac{-1-2x}{3(1+x)^{2+\frac{4}{3}}} =\frac{-1-2x}{3(1+x)^{\frac{10}{3}}}$$ Now we have the derivative of the function: $$\frac{dy}{dx}=\frac{-1-2x}{3(1+x)^{\frac{10}{3}}}$$

Key concepts

Chain Rule Differentiation

The chain rule is a fundamental concept in calculus, employed when differentiating compositions of functions. When we have a function inside another function, like a 'Russian doll', the chain rule enables us to differentiate step by step. In the context of radical functions, where the radical can be considered as an outer function and the inner part as another function, the chain rule becomes particularly useful.

For example, in our exercise, the function inside the radical sign is treated as the inner function 'u', and the radical itself as the outer function. Using the chain rule, we differentiate the outer function with respect to 'u', and then multiply by the derivative of 'u' with respect to 'x'. This method simplifies the process and ensures that all parts of the composition are accounted for in the differentiation process.

Quotient Rule

When differentiating a function that is a ratio of two other functions, we use the quotient rule. This rule states that the derivative of a function \( \frac{f(x)}{g(x)} \) is given by \( \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \). It ensures that the relative rates of change of both the numerator and the denominator are considered.

In the given exercise, we differentiate the inner function 'u', which is a quotient of functions. The application of the quotient rule correlates to the critical step that allows us to find \( \frac{du}{dx} \) correctly before it is integrated into the chain rule. As shown in the step by step solution, using the quotient rule yields a simplified expression which we can then multiply with the derivative obtained from the chain rule to find the final derivative of the original function.

Properties of Exponents

Understanding the properties of exponents is crucial when working with radical functions, especially when they need to be differentiated. Exponents determine how many times a number is multiplied by itself, and radical expressions are just another way of writing expressions with fractional exponents.

In our exercise, the given radical function first needs to be rewritten using exponents to apply differentiation rules properly. Converting \( \sqrt[3]{...} \) to a fractional exponent of \( \frac{1}{3} \) is applying one such property. Recognizing that exponents can be manipulated this way is a pivotal step because it allows us to treat radical expressions just like any other function when differentiating, thus facilitating the use of the chain rule.

Simplifying Derivatives

Simplifying derivatives is the process of reducing the derived expression to its most basic form. This often involves combining like terms, factoring, and canceling common factors. In the case of radical functions, we often end up with complex expressions after applying differentiation rules, and simplifying helps in making the result more interpretable and usable.

The final step in our solved exercise demonstrates simplifying the derivative. After applying the chain rule and quotient rule, and finding the expression for \( \frac{dy}{dx} \), we combine the terms with common bases and add their exponents, according to the properties of exponents. This results in a single, concise derivative that elegantly represents the rate of change of the original radical function.