Problem 77
Question
Differentiate. $$ y=\ln \left(3 x^{2}+2 x-1\right) $$
Step-by-Step Solution
Verified Answer
The derivative is \( \frac{6x + 2}{3x^2 + 2x - 1} \).
1Step 1: Identify the Function Type
The function given is \( y = \ln(3x^2 + 2x - 1) \). This is a composite function where the outer function is the natural logarithm and the inner function is a quadratic expression \( 3x^2 + 2x - 1 \).
2Step 2: Apply the Chain Rule
To differentiate \( y = \ln(f(x)) \), we use the chain rule. The derivative of \( \ln(u) \) with respect to \( u \) is \( \frac{1}{u} \). Therefore, the derivative \( y' \) with respect to \( x \) is \( \frac{1}{f(x)} \cdot f'(x) \).
3Step 3: Differentiate the Inner Function
Differentiate the inner function \( f(x) = 3x^2 + 2x - 1 \). The derivative \( f'(x) \) is \( 6x + 2 \) since the derivative of \( 3x^2 \) is \( 6x \) and the derivative of \( 2x \) is 2.
4Step 4: Combine the Derivatives
Substitute \( f(x) \) and \( f'(x) \) back into the differentiation formula: \[y' = \frac{1}{3x^2 + 2x - 1} \cdot (6x + 2)\]This simplifies to: \[y' = \frac{6x + 2}{3x^2 + 2x - 1}\]
Key Concepts
DifferentiationComposite FunctionsQuadratic Expression
Differentiation
Differentiation is a fundamental concept in calculus. It allows us to find the rate at which a function is changing at any given point. Think of it as understanding how fast a car is going at an exact moment rather than over the course of a trip. This concept is crucial for solving many real-world problems where change is involved—like predicting profits or calculating speed.
In mathematical terms, differentiation gives us a derivative. For a function, the derivative is a new function that tells us the slope of the original function at any point. Finding the derivative often involves applying rules like the chain rule, product rule, or quotient rule. In practice, we perform differentiation on a variety of functions to learn their behavior and predict outcomes.
When you're working with basic functions, differentiation can be as simple as applying familiar formulas. But for more complex situations like composite functions, additional rules like the chain rule become necessary to tackle the problem effectively.
In mathematical terms, differentiation gives us a derivative. For a function, the derivative is a new function that tells us the slope of the original function at any point. Finding the derivative often involves applying rules like the chain rule, product rule, or quotient rule. In practice, we perform differentiation on a variety of functions to learn their behavior and predict outcomes.
When you're working with basic functions, differentiation can be as simple as applying familiar formulas. But for more complex situations like composite functions, additional rules like the chain rule become necessary to tackle the problem effectively.
Composite Functions
Composite functions involve an outer function and an inner function. This means one function exists within another function. These are very common in mathematics, especially in calculus. Imagine a function like nesting dolls, where one is inside the other. The function \( y = \ln(3x^2 + 2x - 1) \) is a perfect example.
Here’s how it breaks down:
By understanding how these functions nest within each other, students can tackle complicated functions efficiently and accurately. It makes breaking down complex problems a lot more effective.
Here’s how it breaks down:
- The outer function is \( \ln(u) \), which is the natural logarithm.
- The inner function is \( 3x^2 + 2x - 1 \), a quadratic expression.
By understanding how these functions nest within each other, students can tackle complicated functions efficiently and accurately. It makes breaking down complex problems a lot more effective.
Quadratic Expression
A quadratic expression is a polynomial that involves a variable raised to the second power. Specifically, it’s of the form \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. Quadratics are everywhere: in physics, engineering, even in the path of a basketball in flight.
In terms of differentiation, finding the derivative of a quadratic expression is straightforward. The power rule applies, where you bring down the exponent and reduce the power by one. For example, the derivative of \( 3x^2 \) becomes \( 6x \), and \( 2x \) becomes \( 2 \). Thus, the derivative of the expression \( 3x^2 + 2x - 1 \) from our example would be \( 6x + 2 \).
Quadratic expressions can form the inner part of composite functions, as we've seen, making it essential to know how to handle them efficiently. Mastering quadratics helps students recognize and solve various problems involving these types of expressions in calculus and other areas of mathematics.
In terms of differentiation, finding the derivative of a quadratic expression is straightforward. The power rule applies, where you bring down the exponent and reduce the power by one. For example, the derivative of \( 3x^2 \) becomes \( 6x \), and \( 2x \) becomes \( 2 \). Thus, the derivative of the expression \( 3x^2 + 2x - 1 \) from our example would be \( 6x + 2 \).
Quadratic expressions can form the inner part of composite functions, as we've seen, making it essential to know how to handle them efficiently. Mastering quadratics helps students recognize and solve various problems involving these types of expressions in calculus and other areas of mathematics.
Other exercises in this chapter
Problem 76
Differentiate. $$ y=\ln \frac{x^{4}}{2} $$
View solution Problem 77
Find an equation of the line tangent to the graph of \(f(x)=e^{2 x}\) at the point (0,1)
View solution Problem 78
Find an equation of the line tangent to the graph of \(G(x)=e^{-x}\) at the point (0,1)
View solution Problem 78
Differentiate. $$ y=\ln \left(7 x^{2}+5 x+2\right) $$
View solution