Problem 53 - Calculus and its applications

Question

Find \(y^{\prime}\) $$ \text { If } y=\sqrt[3]{x}+\sqrt{x}, \text { find }\left.\frac{d y}{d x}\right|_{x=64} $$

Step-by-Step Solution

Answer

The derivative \( y' \) evaluated at \( x = 64 \) is \( \frac{1}{12} \).

Step 1: Understand the problem

We need to find the derivative of the function \( y = \sqrt[3]{x} + \sqrt{x} \) with respect to \( x \), then evaluate it at \( x = 64 \).

Step 2: Apply the power rule

Rewrite the function in terms of exponents: \( y = x^{1/3} + x^{1/2} \). Use the power rule for derivatives: if \( y = x^n \), then \( \frac{dy}{dx} = n x^{n-1} \).

Step 3: Differentiate \(x^{1/3}\)

Using the power rule, \( \frac{d}{dx}(x^{1/3}) = \frac{1}{3}x^{-2/3} \).

Step 4: Differentiate \(x^{1/2}\)

Similarly, apply the power rule: \( \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} \).

Step 5: Combine the derivatives

The derivative of the entire function is the sum of the individual derivatives: \( y' = \frac{1}{3}x^{-2/3} + \frac{1}{2}x^{-1/2} \).

Step 6: Evaluate the derivative at \(x = 64\)

Substitute \( x = 64 \) into the derivative: \( y' = \frac{1}{3}(64)^{-2/3} + \frac{1}{2}(64)^{-1/2} \).

Step 7: Calculate \((64)^{-2/3}\) and \((64)^{-1/2}\)

First, compute: \( 64^{1/3} = 4 \), so \( 64^{-2/3} = \left(\frac{1}{4}\right)^2 = \frac{1}{16} \). For \( 64^{-1/2} \), since \( 64^{1/2} = 8 \), we get \( 64^{-1/2} = \frac{1}{8} \).

Step 8: Final calculation

Substituting these values: \( y' = \frac{1}{3} \times \frac{1}{16} + \frac{1}{2} \times \frac{1}{8} = \frac{1}{48} + \frac{1}{16} \). Convert \( \frac{1}{16} \) to a common denominator: \( \frac{1}{16} = \frac{3}{48} \). Now add: \( \frac{1}{48} + \frac{3}{48} = \frac{4}{48} = \frac{1}{12} \).

Step 9: Conclusion

Thus, the derivative of the given function evaluated at \( x = 64 \) is \( \frac{1}{12} \).

Key Concepts

Power Rule

The power rule is one of the most straightforward and vital tools in calculus for finding the derivative of a function. It is particularly useful when differentiating functions that are simple powers of a variable. If you have a function such as \(y = x^n\), where \(n\) is any real number, the power rule tells us that the derivative of \(y\) with respect to \(x\) is \(\frac{dy}{dx} = nx^{n-1}\). This rule essentially means you multiply the term by its exponent and then decrease the exponent by one.

Let's see how this applies to our exercise. The function \(y = \sqrt[3]{x} + \sqrt{x}\) can be rewritten using exponents as \(y = x^{1/3} + x^{1/2}\). By applying the power rule:

• For \(x^{1/3}\), the derivative is \(\frac{1}{3}x^{-2/3}\).
• For \(x^{1/2}\), the derivative is \(\frac{1}{2}x^{-1/2}\).

Combining these gives the derivative of the entire function.

Differentiation

Differentiation is a method used in calculus to calculate the rate at which a function is changing at any given point. It's the process of finding a derivative, which, in simple terms, is the slope of the function at a particular point. This concept is foundational in understanding the behavior of functions over intervals.

In the context of the given exercise, differentiation was utilized to determine how \(y = \sqrt[3]{x} + \sqrt{x}\) changes concerning \(x\). By rewriting the function in a form that could be easily differentiated using the power rule, we effectively broke down a more complex function into simpler parts. Differentiation allowed us to find that the rate of change for this function is expressed as \(y' = \frac{1}{3}x^{-2/3} + \frac{1}{2}x^{-1/2}\).

This derivative tells us how steep or flat the graph of \(y\) is at any given \(x\). Differentiation helps translate real-world problems into mathematical ones, which can then be solved using calculus.

Evaluating Derivatives at a Point

Evaluating a derivative at a specific point involves substituting a particular value into the derivative function. It allows us to find the slope of the original function at that specific point, providing insight into how rapidly the function is increasing or decreasing.

In this exercise, after finding the derivative \(y' = \frac{1}{3}x^{-2/3} + \frac{1}{2}x^{-1/2}\), the next step was to evaluate it at \(x = 64\). To do so:

• Calculate \((64)^{-2/3}\). Since \(64^{1/3} = 4\), \((64)^{-2/3} = \frac{1}{16}\).
• Calculate \((64)^{-1/2}\). Since \(64^{1/2} = 8\), \((64)^{-1/2} = \frac{1}{8}\).

Substituting these values into the derivative, you perform the arithmetic to obtain \(y' = \frac{1}{12}\). This tells us that at \(x = 64\), the slope of the function \(y = \sqrt[3]{x} + \sqrt{x}\) is \(\frac{1}{12}\), indicating a gentle upward slope at that point.