Question

Differentiate the following functions: \(u=\sqrt[8]{\frac{x}{1-x}}\)

Short answer

Answer: The derivative of the given function is \(u'(x) = -\frac{x^2(\frac{x}{1-x})^{-\frac{7}{8}}}{8(1-x)^2}\).

Step-by-step solution

Rewrite the function

Rewrite the given function as a power function: \(u(x) = (\frac{x}{1-x})^{\frac{1}{8}}\).

Apply the Chain Rule (Differentiate outer function)

Now, applying the chain rule, differentiate \(u(x)\) with respect to \(x\): $$\frac{du}{dx} = \frac{1}{8}(\frac{x}{1-x})^{\frac{1}{8} - 1} \cdot \frac{d}{dx}(\frac{x}{1-x})$$.

Differentiate the inner function

Next, we need to differentiate the inner function \(\frac{x}{1-x}\) with respect to \(x\): $$\frac{d}{dx}(\frac{x}{1-x}) = \frac{(1-x)\frac{d}{dx}(x) - x\frac{d}{dx}(1-x)}{(1-x)^2} = \frac{1-x-x}{(1-x)^2} = \frac{-x^2}{(1-x)^2}$$.

Substitute the derivative of inner function and simplify

Substitute the derivative of \(\frac{x}{1-x}\): $$\frac{du}{dx} = \frac{1}{8}(\frac{x}{1-x})^{\frac{1}{8} - 1} \cdot \frac{-x^2}{(1-x)^2}$$. Simplify the expression: $$\frac{du}{dx} = -\frac{x^2(\frac{x}{1-x})^{-\frac{7}{8}}}{8(1-x)^2}$$. So, the derivative of the given function is: \(u'(x) = -\frac{x^2(\frac{x}{1-x})^{-\frac{7}{8}}}{8(1-x)^2}\).

Key concepts

Chain Rule

The Chain Rule is a fundamental concept in calculus used to differentiate compositions of functions. It's particularly useful when dealing with nested functions where one function is inside another. Here's a simple way to think about it:
The Chain Rule states that to find the derivative of a composite function \( f(g(x)) \), you need to take the derivative of the outer function \( f \) with respect to the inner function \( g \), and then multiply it by the derivative of the inner function \( g \) with respect to \( x \). It can be expressed as:

• \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \)

In the given exercise, we used the Chain Rule to differentiate \( u = \left(\frac{x}{1-x}\right)^{\frac{1}{8}} \). Here, the outer function is \( u = x^{\frac{1}{8}} \) and the inner function is \( g(x) = \frac{x}{1-x} \). This is why the differentiation involved multiplying the derivative of the outer power function by the derivative of the inner rational function.

Power Function

A power function is of the form \( f(x) = x^n \), where \( n \) is any real number. Differentiating power functions is straightforward with the Power Rule, which states:

• \( \frac{d}{dx}[x^n] = n \cdot x^{n-1} \)

This rule applies regardless of whether the exponent is positive, negative, or a fraction. In the original exercise, the given expression \( u = \left(\frac{x}{1-x}\right)^{\frac{1}{8}} \) can be thought of as a power function, \((something)^{\frac{1}{8}}\). The power exponent here is \( \frac{1}{8} \),
and using the Power Rule helps us differentiate the outer portion of the composition. Therefore, the derivative of the outer function is \( \frac{1}{8} \left(\frac{x}{1-x}\right)^{-\frac{7}{8}} \) when applying the Chain Rule.

Fractional Derivatives

Fractional derivatives are an extension of the concept of ordinary differentiation to integration and differentiation of non-integer order. They allow us to apply the idea of taking derivatives any number of times, even a fractional number of times, bringing flexibility in solving complex differential equations.
In the context of our problem, the fractional exponent \( \frac{1}{8} \) in the function \( u = \left(\frac{x}{1-x}\right)^{\frac{1}{8}} \) introduces fractional calculus because the order of differentiation is linked with the order of the power. Applying the Chain Rule and Power Rule together on fractional power functions allows us to handle expressions that are typically more complex than simple integer power functions.
While you may not encounter fractional derivatives in every calculus problem, understanding their basic principles can serve you well in advanced fields like signal processing, system dynamics, and other areas where real-world problems require such nuanced approaches.