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\(37-48\) Use logarithmic differentiation to find the derivative of the function. $$y=(2 x+1)^{5}\left(x^{4}-3\right)^{6}$$

Short Answer

Expert verified
\(\frac{dy}{dx} = 10(2x+1)^4(x^4-3)^6 + 24x^3(2x+1)^5(x^4-3)^5\)

Step by step solution

01

Take the natural logarithm

First, take the natural logarithm of both sides of the equation to simplify the expression. Thus, we have:\[\ln(y) = \ln\left((2x+1)^5(x^4-3)^6\right)\]
02

Apply the logarithmic property

Use the logarithmic property \(\ln(ab) = \ln(a) + \ln(b)\) to expand the right-hand side:\[\ln(y) = \ln((2x+1)^5) + \ln((x^4-3)^6)\]
03

Use the power rule for logarithms

Apply the power rule for logarithms, \(\ln(a^b) = b \ln(a)\), to both terms:\[\ln(y) = 5\ln(2x+1) + 6\ln(x^4-3)\]
04

Differentiate both sides implicitly

Differentiate both sides with respect to \(x\). On the left side, use the chain rule:\[\frac{d}{dx}[\ln(y)] = \frac{1}{y} \frac{dy}{dx}\]On the right side, differentiate each term separately using the chain rule:\[\frac{d}{dx}[5\ln(2x+1)] = \frac{5}{2x+1} \cdot 2 = \frac{10}{2x+1}\]\[\frac{d}{dx}[6\ln(x^4-3)] = \frac{6}{x^4-3} \cdot 4x^3 = \frac{24x^3}{x^4-3}\]
05

Combine derivatives

Combine the derivatives from Step 4:\[\frac{1}{y} \frac{dy}{dx} = \frac{10}{2x+1} + \frac{24x^3}{x^4-3}\]
06

Solve for \(\frac{dy}{dx}\)

Multiply both sides by \(y\) to isolate \(\frac{dy}{dx}\):\[\frac{dy}{dx} = y \left( \frac{10}{2x+1} + \frac{24x^3}{x^4-3} \right)\]Substitute back the original expression for \(y\):\[\frac{dy}{dx} = (2x+1)^5(x^4-3)^6 \left( \frac{10}{2x+1} + \frac{24x^3}{x^4-3} \right)\]
07

Simplify the expression

Distribute and simplify the expression to get the final derivative:\[\frac{dy}{dx} = 10(2x+1)^4(x^4-3)^6 + 24x^3(2x+1)^5(x^4-3)^5\]

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Key Concepts

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

Understanding Derivatives
When studying calculus, one of the most fundamental concepts you'll encounter is the derivative. The derivative represents the rate at which a function is changing at any given point. More simply, it's a way to measure how a function's output changes as its input changes. This concept is crucial for understanding how things change and develop over time in mathematics and in practical scenarios.

For example, if you think of a function as a moving car, the derivative would be the speed at which the car is traveling at any point in time. It tells you how fast or slow the function is progressing. When you take the derivative of a function, you're essentially asking, "How is this function changing?"
  • A positive derivative indicates the function is increasing.
  • A negative derivative indicates the function is decreasing.
  • A zero derivative might indicate a stationary point, such as a peak or a trough.
Exploring the Chain Rule
In calculus, the chain rule is a technique for finding the derivative of a composition of functions. Think of it as a tool to tackle functions within functions—like a series of linked processes that you unravel step-by-step. For example, if you have a function in the form of \(f(g(x))\), the chain rule helps you differentiate it efficiently.

Following the chain rule, the derivative of \(f(g(x))\) is given by \(f'(g(x)) \cdot g'(x)\). This method is essential when dealing with layered functions, and it ensures that you can dissect each part accurately. To apply the chain rule, focus on these steps:
  • Differentiate the outer function. Leave the inside function as it is.
  • Multiply by the derivative of the inner function.
This approach is especially useful in logarithmic differentiation where functions often compound. Applying the chain rule helps break down complex structures into manageable pieces.
The Role of Natural Logarithms
Natural logarithms are logarithms with a base of \(e\), where \(e\) is approximately equal to 2.71828. These logarithms are prevalent in calculus because they simplify differentiation and integration, especially when paired with exponential functions.

In the process of logarithmic differentiation, using the natural logarithm can transform products into sums, quotients into differences, and powers into products. This property makes it easier to handle complicated expressions.
  • For example, \ln(ab) = \ln(a) + \ln(b), which simplifies multiplying terms.
  • Similarly, \ln(a/b) = \ln(a) - \ln(b), assisting in division problems.
  • And for powers, \ln(a^b) = b\ln(a), which breaks down powers for easier manipulation.
Utilizing natural logarithms in differentiation can take an otherwise complex differentiation problem and turn it into a simpler, more manageable task.
Utilizing the Power Rule
The power rule is a fundamental differentiation rule that applies whenever you're dealing with functions of the form \(x^n\). This rule makes it straightforward to differentiate polynomials and simplifies the process for determining rates of change for these functions.

The power rule states that if \(f(x) = x^n\), then the derivative \(f'(x) = n x^{n-1}\). In essence, you bring down the exponent as a coefficient and then reduce the original exponent by one.
  • If \(y = x^3\), then \(f'(x) = 3x^2\).
  • If \(y = x^5\), then \(f'(x) = 5x^4\).
This rule is especially useful when combined with other rules, such as the chain rule and logarithmic differentiation, making it an essential part of the calculus toolkit when dealing with complex functions.

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Most popular questions from this chapter

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