Problem 24 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. \(f(x)=\sqrt[4]{2-4 x^{2}}\)

Step-by-Step Solution

Answer

The derivative is \(-2x(2 - 4x^2)^{-3/4}\).

Step 1: Rewrite the Function

First, express the given function in a form that makes differentiation easier. The original function is written as a fourth root, which can be rewritten using exponents: \[ f(x) = (2 - 4x^2)^{1/4} \]

Step 2: Apply the Chain Rule

To differentiate, we use the chain rule, which states that if you have a composite function \( f(g(x)) \), its derivative is \( f'(g(x)) \cdot g'(x) \). In our case, let \( u = 2 - 4x^2 \), then \( f(x) = u^{1/4} \).

Step 3: Differentiate the Outer Function

Differentiate \( u^{1/4} \) with respect to \( u \). The derivative is: \[ \frac{d}{du}(u^{1/4}) = \frac{1}{4}u^{-3/4} \]

Step 4: Differentiate the Inner Function

Now differentiate the inner function \( u = 2 - 4x^2 \) with respect to \( x \). The derivative is: \[ \frac{d}{dx}(2 - 4x^2) = -8x \]

Step 5: Combine Derivatives Using the Chain Rule

Combine the derivatives from Step 3 and Step 4. According to the chain rule:\[ \frac{d}{dx}f(x) = \frac{1}{4}(2 - 4x^2)^{-3/4} \times (-8x) \]

Step 6: Simplify the Expression

Simplify the expression: \[ \frac{d}{dx}f(x) = -2x(2 - 4x^2)^{-3/4} \]

Key Concepts

Chain Rule

The chain rule is a fundamental differentiation rule, especially important when dealing with composite functions. It's your go-to tool when a function is nested inside another. Imagine a matryoshka doll, where each layer wraps around an inner doll.
In mathematical terms, if you have a function of the form \(f(g(x))\), the chain rule tells you to differentiate with this formula:

• Differentiate the outer function with respect to the inner function.
• Multiply that result by the derivative of the inner function.

This might sound complex, but it's straightforward with practice. Let's see it in action: in the given exercise, \(f(x) = (2 - 4x^2)^{1/4}\). Here, the outer function is \((u)^{1/4}\) and the inner function is \(2 - 4x^2\), with \(u = 2 - 4x^2\).
By applying the chain rule, you differentiate the outer function \((u^{1/4})\) with respect to \(u\), then differentiate the expression \(u\) with respect to \(x\). While it may seem like double the work, each step is smaller and more manageable, breaking down complex tasks into simpler parts.

Composite Functions

Composite functions are functions within functions. They're like an onion with layers; the function \(g(x)\) inside affects the function \(f\) as a whole. Breaking it down, if you read \(f(g(x))\), \(g(x)\) feeds its results into \(f\).
This structure is often seen in calculus, like the example exercise where you have \(f(x) = (2 - 4x^2)^{1/4}\). Here, \((2 - 4x^2)\) is our inside layer or the innermost part, and the fourth root symbolizes the outer layer. Decomposing into these parts makes differentiation easier since you can tackle each layer step by step.
Understanding composite functions allows you to recognize when to apply the chain rule and ensures you’re never lost in nested functions again. It transforms complex problems into straightforward tasks, by addressing each component of the composite function one by one.

Exponent Rules

Even though it might sound daunting, exponent rules are just the toolbox for managing powers in equations. Whether converting roots to fractions or simplifying a power expression, these rules drive simplification in calculus.
In the exercise, the expression \(\sqrt[4]{2-4x^{2}}\) was rephrased as \((2 - 4x^2)^{1/4}\). This is utilizing the rule that \(\sqrt[n]{x}\) can be expressed as \(x^{1/n}\). This transformation is crucial since derivatives are typically easier to handle in exponential form.

• Power Rule: \(a^m \cdot a^n = a^{m+n}\)
• Fractional Exponents: \(\sqrt[n]{a} = a^{1/n}\)
• Negative Exponents: \(a^{-n} = \frac{1}{a^n}\)

These rules are applied initially and during simplification of resulting differentiated terms. By making use of these guidelines, you'll be well-equipped to crack any problem that involves exponents efficiently and straightforwardly.