Question

If \(\mathrm{y}=3 \mathrm{x}^{(3 / 2)}(\mathrm{x}-1)\) then \(|[\mathrm{dy} / \mathrm{d} \mathrm{x}]|_{(\mathrm{x}=1)}=\) (a) 6 (b) 3 (c) 1 (d) 0

Short answer

The absolute value of the derivative of the function \(y=3x^{(3/2)}(x-1)\) evaluated at \(x=1\) is: (b) 3

Step-by-step solution

Identify the function y

The given function is y=3x^(3/2)(x-1).

Apply the product rule

The function can be rewritten as the product of two functions u(x) and v(x): u(x) = 3x^(3/2) v(x) = (x-1) We will now apply the product rule in order to find the derivative. The product rule states that if y = u(x)v(x), then the derivative of y with respect to x is: (dy/dx) = (du/dx)v(x) + u(x)(dv/dx)

Differentiate u(x) and v(x)

First, let's find the derivatives of u(x) and v(x) with respect to x. Apply the power rule to differentiate u(x): du/dx = d(3x^(3/2))/dx = (3/2) * 3x^(1/2) Now, differentiate v(x) with respect to x: dv/dx = d(x-1)/dx = 1

Compute the derivative of y

Now that we have the derivatives of u(x) and v(x), we can compute the derivative dy/dx: dy/dx = (du/dx)v(x) + u(x)(dv/dx) = (3/2) * 3x^(1/2)(x-1) + 3x^(3/2)(1)

Evaluate the derivative at x=1

We need to find the value of dy/dx when x=1: (dy/dx)|_(x=1) = (3/2) * 3(1)^(1/2)(1-1) + 3(1)^(3/2)(1) = 0 + 3

Find the absolute value

Finally, we need to find the absolute value of the derivative evaluated at x=1: |[(dy/dx)|_(x=1)]| = |3| From the provided options, the correct answer is: (b) 3

Key concepts

Understanding the Product Rule

When it comes to differentiating functions that are products of two simpler functions, the product rule is your go-to tool. In simple terms, when you have a function that is made by multiplying two functions, say \( u(x) \) and \( v(x) \), the product rule helps you find the derivative of the combination. The rule states:

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

This formula tells us to take the derivative of the first function and multiply it by the second, then take the derivative of the second function and multiply it by the first.
In our original exercise, we used it to find the derivative of \( y = 3x^{3/2}(x-1) \), which we split into \( u(x) = 3x^{3/2} \) and \( v(x) = x-1 \). The application of the product rule here simplifies finding the derivative by breaking it into simpler steps.

Mastering the Power Rule

The power rule is one of the most straightforward techniques in calculus for finding derivatives. It applies when you have a function of the form \( x^n \), where \( n \) is a constant. The power rule formula is:

• \( \frac{d}{dx}[x^n] = n\cdot x^{n-1} \)

This means that you bring down the power as a coefficient and subtract one from the power itself.
In our exercise, we used the power rule to differentiate \( u(x) = 3x^{3/2} \). Applying the power rule gives:

• \( \frac{du}{dx} = \frac{3}{2} \cdot 3x^{1/2} \)

Understanding and applying the power rule can simplify finding derivatives of functions, especially those involving polynomial expressions. This skill is crucial for solving calculus problems effectively.

The Role of Differentiation

Differentiation is a fundamental concept in calculus, and it's all about finding how a function changes at any given point. Essentially, it's the process of determining the derivative, which represents the rate of change of a function. Differentiation is applied to various fields like physics, engineering, and economics for analyzing dynamic situations.
When differentiating, you're often aiming to find out how one quantity changes in response to changes in another. In our problem, the differentiation process involved applying both the product rule and the power rule to get the derivative of \( y = 3x^{3/2}(x-1) \). This helped us determine the rate at which \( y \) changes as \( x \) changes.
Ultimately, differentiation is about understanding the behavior of functions. It allows us to solve real-world problems by modeling the changes in systems and calculating the points of maximum or minimum values, which is a key technique in various applications.