Question

Differentiate the following functions: \(u=\sqrt[5]{c^{3}-3 c^{2} x+3 c x^{2}-x^{3}}\)

Short answer

Question: Find the derivative of the function \(u = (c^3 - 3c^2x + 3cx^2 - x^3)^{\frac{1}{5}}\) with respect to \(x\). Answer: \(\frac{du}{dx} = \frac{1}{5}(c^3 - 3c^2x + 3cx^2 - x^3)^{-\frac{4}{5}} (-3c^2 + 6cx - 3x^2)\)

Step-by-step solution

Identify the outer and inner functions

The given function can be written as \(u = (c^3 - 3c^2x + 3cx^2 - x^3)^{\frac{1}{5}}\). Here, the outer function is \(f(u) = u^{\frac{1}{5}}\) and the inner function is \(g(x) = c^3 - 3c^2x + 3cx^2 - x^3\).

Differentiate the outer function

Differentiate the outer function with respect to \(u\), we have \(\frac{df}{du} = \frac{1}{5}u^{-\frac{4}{5}}\).

Differentiate the inner function

To differentiate the inner function with respect to \(x\), we get \(\frac{dg}{dx} = -3c^2 + 6cx - 3x^2\).

Apply the chain rule

Using the chain rule, we have \(\frac{du}{dx} = \frac{df}{du}\cdot\frac{dg}{dx}\).

Substitute and simplify

Substituting the derivatives of the outer and inner functions, we get: \(\frac{du}{dx} = \frac{1}{5}u^{-\frac{4}{5}} \cdot (-3c^2 + 6cx - 3x^2) = \frac{1}{5}(c^3 - 3c^2x + 3cx^2 - x^3)^{-\frac{4}{5}} (-3c^2 + 6cx - 3x^2)\). That's the final simplified answer.

Key concepts

Chain Rule Calculus

Understanding the chain rule is crucial for differentiating composite functions where one function is nested inside another. In our exercise, it is applied to find the derivative of a function raised to a fractional power. The chain rule states that to differentiate a composite function, you first differentiate the outer function with respect to its inner function, and multiply that by the derivative of the inner function with respect to the independent variable, represented as \(\frac{du}{dx} = \frac{df}{du} \cdot \frac{dg}{dx}\). This step-by-step approach is especially handy when dealing with polynomials or more complex functions involving exponents, radicals, or trigonometric functions. It ties together the rate of change of the outer function with respect to the inner function and the rate of change of the inner function with respect to the independent variable.

In the provided example, we start by identifying the outer and inner functions and then follow the procedure to differentiate each separately before applying the chain rule. By breaking down the process, we make it more manageable and easier to grasp.

Derivative of Power Functions

Power functions are those in which the variable is raised to a constant power, written in the form \(f(x) = x^n\). The derivative of a power function is found using a basic rule of differentiation: if \(n\) is a real number, then the derivative of \(x^n\) with respect to \(x\) is \(nx^{n-1}\). In our example, the power function is slightly more complex because it involves a fifth root, which is represented as a power of \(\frac{1}{5}\). Applying the power rule, we derive the outer function \(u^{\frac{1}{5}}\) to get \(\frac{1}{5}u^{-\frac{4}{5}}\).

This rule simplifies the process of finding derivatives and is one of the first tools students learn in calculus for handling polynomial functions. By understanding the derivative of power functions, we lay the groundwork for tackling a wide variety of problems involving rates of change and function behavior.

Implicit Differentiation

While implicit differentiation was not directly used in the exercise, it's worthwhile to understand its role in calculus. Implicit differentiation becomes necessary when dealing with equations where the dependent and independent variables are interwoven, and it's not feasible to explicitly solve for one in terms of the other. This technique allows us to find the derivative without directly isolating the variable.

By treating every occurrence of \(y\) as a function of \(x\) and applying the chain rule, we differentiate implicitly. Let's consider the example \(x^2 + y^2 = 1\), where solving for \(y\) explicitly might be tedious. Instead, by differentiating both sides with respect to \(x\) and applying implicit differentiation, we readily obtain \(\frac{dy}{dx}\) as a function of \(x\) and \(y\). It’s a technique that broadens our ability to handle more intricate equations, especially those arising in physics and engineering.