Problem 4 - Thomas Calculus

Question

Find \(d y / d x\) in Exercises \(1-10\) $$ y=\sqrt[4]{5 x} $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{1}{4 \sqrt[4]{(5x)^3}} \)

Step 1: Rewrite the Function Using Exponents

The original function given is \( y = \sqrt[4]{5x} \). To make differentiation easier, rewrite this function using exponents: \( y = (5x)^{1/4} \).

Step 2: Apply the Chain Rule

To differentiate \( y = (5x)^{1/4} \), apply the chain rule. The chain rule states that if you have a composite function \( y = u^n \), then \( \frac{dy}{dx} = n u^{n-1} \cdot \frac{du}{dx} \). Here, \( u = 5x \) and \( n = 1/4 \).

Step 3: Differentiate the Outer Function

Differentiate the outer part of the function, treating \( 5x \) as a constant factor. Using \( y = u^{1/4} \), we find \( \frac{dy}{du} = \frac{1}{4}u^{-3/4} \).

Step 4: Differentiate the Inner Function

Now, differentiate the inner function \( u = 5x \). The derivative of \( 5x \) with respect to \( x \) is \( \frac{du}{dx} = 5 \).

Step 5: Combine the Results from the Chain Rule

Combine the derivatives: \[ \frac{dy}{dx} = \frac{1}{4}(5x)^{-3/4} \times 5 = \frac{5}{4}(5x)^{-3/4} \].

Step 6: Simplify the Expression

Simplify the expression: \[ \frac{dy}{dx} = \frac{5}{4} \cdot \frac{1}{(5x)^{3/4}} = \frac{5}{4 \cdot (5x)^{3/4}} \]. After simplification, \[ \frac{dy}{dx} = \frac{1}{4 \sqrt[4]{(5x)^3}} \].

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the rate at which a function changes at any point. It's like figuring out the slope of a function at a specific point. Think of it as finding the steepness of a curve, just as you would measure the steepness of a hill.
Differentiating a function means you are calculating its derivative. This derivative tells us how fast the function's value is changing at any given moment.

• To differentiate a simple function like \( y = 5x \), you would find that the derivative is just \( 5 \).
• For more complex functions, especially those that are products or compositions of other functions, specific rules like the chain rule (which we'll cover soon) are needed.
• The notation \( dy/dx \) is commonly used to denote a derivative, highlighting the change in \( y \) with respect to a change in \( x \).

Differentiation is useful in many fields, from physics to economics. It helps in understanding phenomena like speed, acceleration, and even the growth rate of investments.

Chain Rule

The chain rule is a critical tool in calculus for differentiating composite functions (functions composed of other functions). It provides a systematic way of finding the derivative of a function when a function is nested within another.
Consider a simple example of a composite function, such as \( y = \sqrt[4]{5x} \). Here:

• The outer function is the root function \( u^{1/4} \).
• The inner function is \( 5x \).

The chain rule states:\[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\]This means:

• First, differentiate the outer function, treating the inner function as its variable.
• Then, multiply by the derivative of the inner function.

The chain rule makes understanding the mechanics of more complex functions much simpler. It might seem tricky at first, but with practice, it becomes a powerful tool in your calculus toolkit.

Exponents

Exponents, also known as powers, are a shorthand way of expressing repeated multiplication of a number by itself. For example, \( x^3 \) means \( x \times x \times x \). Exponents are also used in calculus to transform expressions, making differentiation straightforward.
When dealing with functions underlined with roots, like \( y = \sqrt[4]{5x} \), recalibrate them using fractional exponents for easy differentiation. This example can be rewritten as \( y = (5x)^{1/4} \).
Key points when working with exponents:

• Fractional exponents denote roots. For instance, \( x^{1/2} \) represents the square root of \( x \), while \( x^{1/4} \) implies the fourth root.
• Negative exponents indicate division instead of multiplication. \( x^{-1} \) equates to \( 1/x \).
• When differentiating, the power rule \( x^n \rightarrow nx^{n-1} \) is frequently employed.

Becoming comfortable with these transformations is crucial for solving calculus problems efficiently, as it clarifies how to handle roots and powers in equations.

Simplification

Simplification is the process of making an expression or equation easier to read and work with by reducing its complexity. It's an essential skill in mathematics that allows one to focus on honing calculations.
After using calculus rules like the chain rule, you'll often have an intermediate result that needs to be further simplified. For instance, after differentiating \( y = (5x)^{1/4} \), you get a complex expression. Simplifying boils down to reducing this expression to its most basic form without losing its essence.

• This involves simplifying fractions or radicals. Taking \( \frac{5}{4 \cdot (5x)^{3/4}} \) to \( \frac{1}{4 \sqrt[4]{(5x)^3}} \) is a classic example.
• Simplification is crucial to make the results interpretable and usable, especially in practical applications.

A simpler form often means better clarity and ease of communication, both in academic settings and real-world scenarios.

Composite Function

A composite function is essentially a function within a function. Understanding how to recognize and work with composite functions is fundamental in calculus.
For example, the function \( y = \sqrt[4]{5x} \) is composite because you have two operations happening: multiplying \( x \) by \( 5 \), and then taking the fourth root. Neither operation stands alone; they combine to define the overall function.

• Composite functions are typically noted with \( (f \circ g)(x) = f(g(x)) \). Here, \( g(x) \) is the inner function and \( f(x) \) is the outer function.
• The process of breaking down a composite function into its components is crucial for applying calculus techniques like the chain rule.
• Recognizing composite functions aids in determining the sequence of operations and allows for predictable manipulation and evaluation.

Mastering composite functions, you can tackle more sophisticated problems, enhancing both mathematical intuition and analytical skills.