Chapter 3: Problem 17
Find \(D_{x} y\). $$ y=(3 x-2)^{2}\left(3-x^{2}\right)^{2} $$
Short Answer
Expert verified
Use the product rule and simplify: \( D_x y = 6(3x-2)(3-x^2)^2 - 4x(3x-2)^2(3-x^2) \).
Step by step solution
01
Identify the Function Structure
The function is a product of two parts: \((3x-2)^2\) and \((3-x^2)^2\). We will use the product rule to differentiate it.
02
Apply the Product Rule
According to the product rule, if \(y = u imes v\), then \(D_x y = u'v + uv'\). Here, \(u = (3x-2)^2\) and \(v = (3-x^2)^2\).
03
Differentiate the First Part (u)
Set \(u = (3x-2)^2\). Differentiate \(u\) with respect to \(x\) using the chain rule: \(u' = 2(3x-2) \cdot 3 = 6(3x-2)\).
04
Differentiate the Second Part (v)
Set \(v = (3-x^2)^2\). Differentiate \(v\) with respect to \(x\) using the chain rule: \(v' = 2(3-x^2) \cdot (-2x) = -4x(3-x^2)\).
05
Substitute into the Product Rule Formula
Substitute \(u\), \(v\), \(u'\), and \(v'\) into the product rule formula: \(D_x y = [6(3x-2)](3-x^2)^2 + (3x-2)^2[-4x(3-x^2)]\).
06
Simplify the Expression
Distribute and combine like terms in the expression obtained from the product rule to simplify: First Term: \(6(3x-2)(3-x^2)^2\) Second Term: \(-4x(3x-2)^2(3-x^2)\). Expand and simplify both terms, then combine them carefully to get the final expression for \(D_x y\).
07
Final Answer
After simplification, the derivative \(D_x y\) is: \[ D_x y = 6(3x-2)(3-x^2)^2 - 4x(3x-2)^2(3-x^2) \]. Further factor if possible.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule
In calculus, the chain rule is a fundamental technique for differentiating composite functions. It helps you find the derivative of a function that is nested within another function. Consider a function of the form \( h(x) = f(g(x)) \). Here, the function \( g(x) \) is inside the function \( f \), making it a composite function.
The chain rule states that you differentiate \( h(x) \) with respect to \( x \) by multiplying the derivative of the outer function \( f \) evaluated at \( g(x) \) by the derivative of the inner function \( g(x) \). The formula is:
The chain rule states that you differentiate \( h(x) \) with respect to \( x \) by multiplying the derivative of the outer function \( f \) evaluated at \( g(x) \) by the derivative of the inner function \( g(x) \). The formula is:
- \( h'(x) = f'(g(x)) \cdot g'(x) \)
Differentiation
Differentiation is the process used in calculus to compute the derivative of a function. It measures how a function changes as its input changes, giving us the function's rate of change or slope at any point. In practical terms, the derivative tells us how steep a curve is at any specific point along the graph of a function.
When applying differentiation, we often utilize various rules:
When applying differentiation, we often utilize various rules:
- Constant Rule: The derivative of a constant is zero.
- Power Rule: For \( x^n \), the derivative is \( nx^{n-1} \).
- Sum Rule: The derivative of a sum of functions is the sum of their derivatives.
- Product Rule: Used when dealing with the product of two functions, described in detail below.
Derivatives
Derivatives represent a cornerstone concept in calculus, symbolizing the instantaneous rate of change of a function with respect to one of its variables. Derivatives are not just limited to simple expressions but apply across polynomials, trigonometric, exponential functions, and beyond.
The notation \( f'(x) \) or \( \frac{df}{dx} \) is often used to denote the derivative of a function \( f \). The idea of taking a derivative is derived from the concept of a limit, which allows us to see how a function behaves as it approaches a certain point.
Understanding derivatives is critical for:
The notation \( f'(x) \) or \( \frac{df}{dx} \) is often used to denote the derivative of a function \( f \). The idea of taking a derivative is derived from the concept of a limit, which allows us to see how a function behaves as it approaches a certain point.
Understanding derivatives is critical for:
- Determining the slope at a given point on a graph.
- Identifying local extrema (i.e., maximums and minimums).
- Analysing real-world scenarios, like velocity in physics or marginal cost in economics.
Function Structure
The structure of a function fundamentally affects the approach used to differentiate it. When encountering a function that is the product of two or more expressions, examining its structure helps us apply the correct differentiation rules.
In multicomponent functions like those presented in the exercise, identifying the function's constituents is essential:
In multicomponent functions like those presented in the exercise, identifying the function's constituents is essential:
- Identify each part of the function, for example, \( u \) and \( v \).
- Utilize appropriate differentiation rules, e.g., product rule or chain rule, as required.
- Re-check the structure, ensuring correct application of the relevant derivative rules and simplification method.