Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-20, find \(D_{x} y\). $$ y=\frac{1}{\left(3 x^{2}+x-3\right)^{9}} $$

Step-by-Step Solution

Answer

The derivative \(D_x y\) is \(\frac{-9(6x + 1)}{(3x^2 + x - 3)^{10}}\).

Step 1: Identify the Rule

To differentiate the function \(y = \frac{1}{(3x^2 + x - 3)^9}\), recognize it requires the use of the chain rule. The chain rule states \( \frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x) \). Here, consider \(u = (3x^2 + x - 3)\) and \(y = u^{-9}\).

Step 2: Differentiate the Outer Function

Apply the power rule to differentiate the outer function. This gives \(\frac{d}{du}(u^{-9}) = -9u^{-10}\).

Step 3: Differentiate the Inner Function

Differentiate \(u = 3x^2 + x - 3\) with respect to \(x\). Applying the sum rule and power rule, we get \(u' = 6x + 1\).

Step 4: Apply the Chain Rule

Combine the results from steps 2 and 3. Substitute back the expressions to find \(\frac{dy}{dx} = -9(3x^2 + x - 3)^{-10} \cdot (6x + 1)\).

Step 5: Simplify the Derivative

Simplify the expression obtained in Step 4. The derivative becomes \(D_x y = \frac{-9(6x + 1)}{(3x^2 + x - 3)^{10}}\).

Key Concepts

Understanding the Chain Rule

The chain rule is a fundamental concept in calculus differentiation. It's used when you have functions nested within others, also known as composite functions. For instance, when a function is in the form of \( f(g(x)) \), the chain rule helps differentiate it by focusing on one layer at a time. The standard formula of the chain rule is:

• \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \)

Let's break it down. Consider a situation where you have \( y = \frac{1}{(3x^2 + x - 3)^9} \). Here, identify the outer function and inner function:

• Outer function: \( u^{-9} \)
• Inner function: \( u = 3x^2 + x - 3 \)

The purpose is to differentiate the outer function with respect to the inner function (using \( u \)), and then differentiate the inner function \( u \) with respect to \( x \). Multiply these derivatives to obtain the derivative of the composite function with respect to \( x \).
This systematic approach might initially seem complex, but by understanding the idea of nesting, applying the chain rule becomes intuitive. Especially when targeting composite functions, this rule is an invaluable tool, making the seemingly complex simpler to navigate.

Applying the Power Rule

The power rule is another essential technique in differentiation which makes dealing with exponents straightforward. It states that if you have a function \( y = x^n \), then its derivative is found using the formula:

• \( \frac{d}{dx}x^n = nx^{n-1} \)

Imagine you need to apply the power rule to a function like \( y = u^{-9} \). Here, the power rule entails lowering the exponent \(-9\) by one and multiplying by the original exponent:

• Derivative: \( -9u^{-10} \)

Once you use the power rule on the outer function, it becomes simpler to calculate the derivative of the entire expression. As we've separated the calculation of \( u \) previously, the power rule application solely focuses on the power and exponent part of the function. This method is not only efficient but also reduces the chance of errors associated with manually expanding and distributing terms. Once familiar, its utility in calculus becomes second nature.

Simplifying the Derivative

Simplifying derivatives is often required to present an answer in its neatest form. After deriving using the chain and power rule, expressions can become bulky, with multiple terms. Simplification helps make these expressions both easier to interpret and more practical for evaluating at given points.In our exercise, we found:

• Derivative: \( \frac{dy}{dx} = -9(3x^2 + x - 3)^{-10} \cdot (6x + 1) \)

To simplify, you apply algebraic techniques, consolidating terms and reducing the expression to its simplest form:

• \( D_x y = \frac{-9(6x + 1)}{(3x^2 + x - 3)^{10}} \)

This achieved through steps such as combining coefficients or rewriting terms in a more compact form.
Simplification not only makes the derivative cleaner but also essential when further analysis or calculation is needed. A simplified derivative is easier to work with in subsequent integration or calculation tasks. Learning how to adeptly simplify expressions is a vital skill for efficiency and clarity in calculus.