Problem 61

Question

Differentiate with respect to the independent variable. $f(x)=\sqrt{3 x}\left(x^{2}-1\right)$

Step-by-Step Solution

Verified

Answer

The derivative is $ f'(x) = \frac{3(x^2 - 1)}{2\sqrt{3x}} + 2x\sqrt{3x} $.

1Step 1: Apply the Product Rule

The given function is in the form of a product: $ f(x) = u(x) imes v(x) $ where $ u(x) = \sqrt{3x} $ and $ v(x) = x^2 - 1 $. According to the product rule, the derivative $ f'(x) $ is given by $ f'(x) = u'(x)v(x) + u(x)v'(x) $. We will find $ u'(x) $ and $ v'(x) $ in the next steps.

2Step 2: Differentiate $ u(x) = \sqrt{3x} $

Rewrite $ \sqrt{3x} $ as $ (3x)^{1/2} $. Using the chain rule, find the derivative: $ u'(x) = \frac{1}{2}(3x)^{-1/2} \times 3 = \frac{3}{2\sqrt{3x}} $.

3Step 3: Differentiate $ v(x) = x^2 - 1 $

Find the derivative of $ v(x) $: $ v'(x) = 2x $.

4Step 4: Compute $ f'(x) $ using the Product Rule

Substitute $ u'(x) $, $ v(x) $, $ u(x) $, and $ v'(x) $ into the product rule formula: $ f'(x) = \frac{3}{2\sqrt{3x}}(x^2 - 1) + \sqrt{3x}(2x) $.

5Step 5: Simplify the Expression

First, simplify $ \frac{3}{2\sqrt{3x}}(x^2 - 1) $ to $ \frac{3(x^2 - 1)}{2\sqrt{3x}} $. Then, simplify $ \sqrt{3x}(2x) $ to $ 2x\sqrt{3x} $. Combine the two terms for $ f'(x) $: $ f'(x) = \frac{3(x^2 - 1)}{2\sqrt{3x}} + 2x\sqrt{3x} $.

6Step 6: Further Simplify if Possible

Combine under a common denominator if possible. Here, it is not necessary to find a common denominator since they are distinct. The derivative $ f'(x) = \frac{3(x^2 - 1)}{2\sqrt{3x}} + 2x\sqrt{3x} $ is our final simplified form.

Key Concepts

Understanding the Product RuleExploring the Chain RuleGrasping Differentiation Basics

Understanding the Product Rule

The product rule is a fundamental tool in calculus used when differentiating functions that are products of two other functions. If you have a function like our example, $ f(x) = u(x) \times v(x) $, where $ u(x) $ and $ v(x) $ are both functions of $ x $, the product rule tells us how to take the derivative. By applying the product rule, we express the derivative as:

$ f'(x) = u'(x)v(x) + u(x)v'(x) $

This formula helps us determine how the rate of change of the whole product function relates to the rates of change of its factors.
If one factor changes, this affects the entire product function, which is why we need to consider both $ u'(x) $ and $ v'(x) $ in the rule.
In practice, remember:

First, differentiate $ u(x) $ and $ v(x) $ separately,
Then plug them into the product rule formula.

This method allows us to effectively differentiate functions represented as products of simpler functions.

Exploring the Chain Rule

The chain rule is a differentiation technique used to handle composite functions, those where one function is applied to the result of another. It's essential when dealing with functions within functions. For example, taking the derivative of $ u(x) = \sqrt{3x} $ involves using the chain rule because you have the square root function applied to $ 3x $.
To differentiate using the chain rule, you:

Consider the outer function and find its derivative as if everything inside it is a single variable.
Then multiply by the derivative of the inside function.

For $ u(x) = (3x)^{1/2} $, the outer function is $ x^{1/2} $, and the inner function is $ 3x $.
The chain rule steps are:

Differentiate the outer function: $ \frac{1}{2}(3x)^{-1/2} $.
Find the derivative of the inner function: $ 3 $.
Multiply the two results: $ \frac{3}{2\sqrt{3x}} $.

Using the chain rule simplifies the differentiation of complex, nested functions into manageable parts.

Grasping Differentiation Basics

Differentiation is the process of finding a derivative, which measures how a function changes as its input changes. It's a key concept in calculus and reflects various real-world rates, such as speed.
To differentiate a given function systematically:

Identify the basic derivative rules, like for power functions $ x^n $, which is $ nx^{n-1} $.
Know how to handle combinations of functions, such as products and compositions, by applying rules like the product rule and chain rule.

Differentiation helps us determine instantaneous rates of change. In our exercise, we used both the product and chain rule to find the derivative.
This shows how differentiation combines basic rules and special techniques to address complex expressions. Understanding differentiation is crucial for deeper insight into the behavior and characteristics of functions.

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