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Chapter 2: Problem 39

Find \(d y / d x\) $$ y=\frac{(2 x+3)^{3}}{\left(4 x^{2}-1\right)^{8}} $$

Short Answer

Expert verified
Apply the quotient and chain rules to differentiate: \(d y/d x = \frac{(4x^2-1)^8 (6(2x+3)^2) - (2x+3)^3 (64x(4x^2-1)^7)}{((4x^2-1)^8)^2}\)."

Step by step solution

01

Identify the Differentiation Rule

The function given is a quotient of two functions, so we need to use the quotient rule for differentiation, which states that if you have a function given by \(\frac{u}{v}\), then its derivative is \(\frac{v \cdot u' - u \cdot v'}{v^2}\). Here \(u=(2x+3)^3\) and \(v=(4x^2-1)^8\).
02

Differentiate the Numerator Using Chain Rule

To find \(u'\), we use the chain rule on \(u=(2x+3)^3\). Let \(g(x) = 2x+3\), then \(u = g(x)^3\). The derivative using chain rule is \(u' = 3g(x)^2 \cdot g'(x) = 3(2x+3)^2 \cdot 2 = 6(2x+3)^2\).
03

Differentiate the Denominator Using Chain Rule

To find \(v'\), apply the chain rule to \(v=(4x^2-1)^8\). Let \(f(x) = 4x^2-1\), then \(v = f(x)^8\). Using the chain rule, \(v' = 8f(x)^7 \cdot f'(x) = 8(4x^2-1)^7 \cdot 8x = 64x(4x^2-1)^7\).
04

Apply the Quotient Rule

Substitute \(u\), \(u'\), \(v\), and \(v'\) into the quotient rule formula: \[\frac{d y}{d x} = \frac{(4x^2-1)^8 \cdot 6(2x+3)^2 - (2x+3)^3 \cdot 64x(4x^2-1)^7}{((4x^2-1)^8)^2}\] Simplify the expression wherever possible, which will give the derivative of the original function.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quotient Rule
The Quotient Rule is a fundamental technique in calculus used for finding the derivative of a quotient of two functions. When you have a function that is divided by another, such as \( \frac{u}{v} \), the quotient rule is your go-to method. It is essential because it helps to efficiently compute derivatives in situations where one function is nested within another.
Here's how it works:
  • Consider \( u = (2x+3)^3 \) and \( v = (4x^2-1)^8 \).
  • The quotient rule states that, \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2} \).
  • This means we need to find the derivatives of both \( u \) and \( v \), represented by \( u' \) and \( v' \) respectively.
Breaking down the derivation according to this rule allows us to find the derivative step by step while ensuring accuracy.
Chain Rule
The Chain Rule is another key tool in the differentiator's toolkit, particularly useful when dealing with composite functions鈥攆unctions within other functions. Imagine peeling an onion layer by layer; the chain rule applies this concept to function differentiation.
For our problem:
  • We need to apply the chain rule to find \( u' \) and \( v' \).
  • For \( u = (2x+3)^3 \), we let \( g(x) = 2x+3 \), thus \( u = (g(x))^3 \).
  • So \( u' = 3(g(x))^2 \cdot g'(x) = 6(2x+3)^2 \).
  • For \( v = (4x^2-1)^8 \), insights are drawn by letting \( f(x) = 4x^2-1 \), hence \( v = (f(x))^8 \).
  • Thus, \( v' = 8(f(x))^7 \cdot f'(x) = 64x(4x^2-1)^7 \).
This step-wise application of the chain rule simplifies the differentiation process, cutting through complex compositions with ease.
Derivative Simplification
After applying derivative techniques like the quotient and chain rules, it's crucial to simplify the result. Simplifying derivative expressions not only makes the solution clearer, but can also reduce computational workload in subsequent calculations.
Following the differentiation of both the numerator and the denominator, their expressions are then incorporated into the quotient rule formula:
  • Substitute \(u = (2x+3)^3\), \(u' = 6(2x+3)^2\), \(v = (4x^2-1)^8\), and \( v' = 64x(4x^2-1)^7\).
  • The formula becomes \[ \frac{(4x^2-1)^8 \cdot 6(2x+3)^2 - (2x+3)^3 \cdot 64x(4x^2-1)^7}{((4x^2-1)^8)^2} \]
  • Now, simplify where possible. This might involve expanding terms or factoring to make the solution more understandable and manageable.
Each simplification step makes the final derivative expression more elegant and easier to interpret.

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Most popular questions from this chapter

Given that \(f^{\prime}(x)=\frac{x}{x^{2}+1}\) and \(g(x)=\sqrt{3 x-1},\) find \(F^{\prime}(x)\) if \(F(x)=f(g(x))\)

Find \(d y / d x\) $$ y=\left(x^{2}+x\right)^{5} \sin ^{8} x $$

Given the following table of values, find the indicated derivatives in parts (a) and (b). $$ \begin{array}{|c|c|c|c|c|}\hline x & {f(x)} & {f^{\prime}(x)} & {g(x)} & {g^{\prime}(x)} \\ \hline-1 & {2} & {3} & {2} & {-3} \\ \hline 2 & {0} & {4} & {1} & {-5} \\ \hline\end{array} $$ (a) \(F^{\prime}(-1),\) where \(F(x)=f(g(x))\) (b) \(G^{\prime}(-1),\) where \(G(x)=g(f(x))\)

Find the indicated derivative. $$ \lambda=\left(\frac{a u+b}{c u+d}\right)^{6} ; \text { find } \frac{d \lambda}{d u} \quad(a, b, c, d \text { constants }) $$

You are asked in these exercises to determine whether a piecewise-defined function \(f\) is differentiable at a value \(x=x_{0}\) where \(f\) is defined by different formulas on different sides of \(x_{0} .\) You may use without proof the following result, which is a consequence of the Mean-Value Theorem (discussed in Section \(4.8) .\) Theorem. Let \(f\) be continuous at \(x_{0}\) and suppose that \(\lim _{x \rightarrow x_{0}} f^{\prime}(x)\) exists. Then \(f\) is differentiable at \(x_{0},\) and \(f^{\prime}\left(x_{0}\right)=\lim _{x \rightarrow x_{0}} f^{\prime}(x) .\) $$ \begin{array}{l}{\text { Let } \quad f(x)=\left\\{\begin{array}{ll}{x^{2},} & {x \leq 1} \\ {\sqrt{x},} & {x>1}\end{array}\right.} \\ {\text { Determine whether } f \text { is differentiable at } x=1 . \text { If so, find }} \\\ {\text { the value of the derivative there. }}\end{array} $$
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