Problem 32 - Calculus an Applied Approach

Question

Find the value of the derivative of the function at the given point. $$ {y=3 x\left(x^{2}-\frac{2}{x}\right)} \quad (2,18) $$

Step-by-Step Solution

Answer

The value of the derivative of the function at the point (2,18) is \(27.5\).

Step 1: Apply the Product Rule

Apply the product rule for derivatives to the function \(y\). For \(y=3x\left(x^2-\frac{2}{x}\right)\), the product rule gives, \(\frac{dy}{dx} = 3x*\frac{d}{dx}\left(x^2-\frac{2}{x}\right) + \left(x^2-\frac{2}{x}\right)*\frac{d}{dx}(3x)\).

Step 2: Calculate the derivatives

The derivative of \(x^2-\frac{2}{x}\) is \(2x+\frac{2}{x^2}\)and the derivative of \(3x\) is 3. Substituting these into the product rule gives, \(\frac{dy}{dx} = 3x*(2x+\frac{2}{x^2}) + (x^2-\frac{2}{x})*3\).

Step 3: Simplify the equation

After simplification, this gives us \(\frac{dy}{dx}=6x^2+6-\frac{6}{x}\).

Step 4: Evaluate the derivative at the given point

Finally, substitute \(x=2\) into the derivative to find its value at the given point. This gives \(\frac{dy}{dx} = 6*2^2+6-\frac{6}{2} = 24+\frac{3}{2}\).

Key Concepts

Product Rule

When we deal with multiplying two or more functions, calculating derivatives can get complex.
The Product Rule is a helpful tool for differentiation in such scenarios. It states that the derivative of a product of two functions, say, \( u(x) \) and \( v(x) \), is given by:

• \( \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \)

This formula allows you to find a derivative by differentiating each function separately and then combining the results.
In the original exercise, we use the Product Rule to separate the differentiation of \( 3x \) from \( x^2 - \frac{2}{x} \). By applying it step-by-step, you can easily dissect and solve more challenging calculus problems. Remember, practice makes perfect when it comes to mastering the Product Rule!

Differentiation

Differentiation is a fundamental concept in calculus used to find the rate at which a function is changing at any given point. This is represented mathematically as the derivative. The process involves applying rules that tell us how to differentiate different kinds of functions.
In our exercise, the differentiation needed two separate operations:

• Calculating the derivative of \( x^2 - \frac{2}{x} \)
• And the derivative of \( 3x \)

The power rule is applied to differentiate \( x^2 \), resulting in \( 2x \).
Meanwhile, the function \( -\frac{2}{x} \) can be rewritten as \( -2x^{-1} \), and its derivative is \( \frac{2}{x^2} \).
The derivative of \( 3x \) is simply 3 because it is linear.
This systematic approach to differentiation helps simplify seemingly complex functions into more manageable parts.

Function Evaluation

Once we've calculated the derivative, evaluating it at specific points is crucial to understand its behavior at those points.
This is termed "Function Evaluation." In the context of the given exercise, after finding the derivative \( \frac{dy}{dx} = 6x^2 + 6 - \frac{6}{x} \), we substitute \( x = 2 \) to find the slope of the tangent to the function at that point. Replacing \( x \) with 2, we calculate:

• \( 6*2^2 = 24 \)
• \( 6 - \frac{6}{2} = 3 \)

Adding together, the value at \( x = 2 \) becomes \( 24 + 3 = 27 \).
This tells us precisely how fast the function \( y = 3x(x^2 - \frac{2}{x}) \) is changing when \( x = 2 \).
Understanding function evaluation helps interpret derivative results in practical terms, allowing you to see how changes in input reflect as changes in output.