Question

Find \(d y / d x\). \(y=\sqrt[3]{x}-2 x^{7 / 2}\)

Short answer

The derivative is \( \frac{1}{3}x^{-2/3} - 7x^{5/2} \).

Step-by-step solution

Differentiate the First Term

The first term of the function is \( \sqrt[3]{x} \), which can be rewritten as \( x^{1/3} \). To find the derivative of this term with respect to \( x \), use the power rule: \( \frac{d}{dx} x^n = n x^{n-1} \). Applying this rule, the derivative is \( \frac{1}{3}x^{-2/3} \).

Differentiate the Second Term

The second term of the function is \( -2 x^{7/2} \). Use the power rule to find its derivative: \( \frac{d}{dx} x^n = n x^{n-1} \). Multiply by the constant \(-2\), so the derivative is \(-2 \cdot \frac{7}{2} x^{7/2-1} = -7x^{5/2} \).

Combine the Derivatives

Combine the derivatives of the two terms from Steps 1 and 2 to get the derivative of the entire function. The derivative \( \frac{dy}{dx} \) is \( \frac{1}{3}x^{-2/3} - 7x^{5/2} \).

Key concepts

Power Rule

The power rule is a key tool in calculus for finding derivatives of polynomial expressions. It's a simple and powerful technique, making it an essential part of differentiation. The basic formula for the power rule is \( \frac{d}{dx} x^n = n x^{n-1} \). This means that to differentiate \( x^n \), you bring the exponent \( n \) down in front of the expression as a coefficient, and then reduce the exponent by one.
For example, in the original exercise, the term \( \sqrt[3]{x} \) is rewritten as \( x^{1/3} \). Applying the power rule here, the derivative becomes \( \frac{1}{3} x^{-2/3} \). Notice how we multiplied the exponent, \( 1/3 \), by the original term and decreased the exponent by one.
Similarly, when differentiating the term \( -2 x^{7/2} \) using the power rule:

• The exponent \( 7/2 \) is brought down as part of the coefficient.
• Multiply to get \( -2 \times 7/2 = -7 \).
• Finally, the exponent reduces by one, giving \( x^{5/2} \).

This results in the derivative \( -7x^{5/2} \). Understanding the power rule helps simplify the differentiation process significantly. This makes it a foundational concept for solving calculus problems.

Differentiation Techniques

Differentiation involves various techniques depending on the type of function you're dealing with. In the example provided, we've used the power rule, but it's important to be aware of other methods too. Different techniques might come into play based on the complexity of the functions. Besides the power rule, here are a few other common differentiation methods:

• Product Rule: Used when you're differentiating a product of two functions.
• Quotient Rule: Applies when dividing two functions.
• Chain Rule: Necessary for differentiating composite functions.

Understanding when and how to apply these techniques is vital. They extend the basic application of the power rule to more complex expressions.
For example, should there have been a composite function in the original exercise, like \( (3x^2 + 1)^4 \), we would use the chain rule to tackle the outer and inner functions separately.
By mastering these differentiation techniques, you will enhance your ability to solve varied calculus problems systematically.

Calculus Problem Solving

Approaching calculus problems systematically involves recognizing the mathematical operations required and applying the correct differentiation techniques. In the original exercise, the aim was to find the derivative \( \frac{dy}{dx} \) of the function \( y = \sqrt[3]{x} - 2x^{7/2} \).
The first step involves transforming the expression into a form that's easier to differentiate. Often, this means rewriting radical expressions as exponents, as done with \( \sqrt[3]{x} \) into \( x^{1/3} \).
After rewriting, you apply the differentiation rules:

• Use the power rule for each term separately.
• Combine the results to form the derivative of the entire function.

By doing so, you arrive at the derivative: \( \frac{1}{3}x^{-2/3} - 7x^{5/2} \).
Being methodical in your approach minimizes errors and ensures a clear path to the solution. Besides calculations, understanding the "why" behind each differentiation technique builds your confidence in solving diverse calculus problems.
Remember, problem-solving in calculus is not just about reaching the answer, but about understanding the process and the mechanics of the functions you're analyzing.