/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 Differentiate the function. $$... [FREE SOLUTION] | 91Ó°ÊÓ

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Differentiate the function. $$ f(x)=\ln (2 x+3) $$

Short Answer

Expert verified
The derivative of the given function, \(f(x) = \ln(2x + 3)\) is \(f'(x) = \frac{2}{2x+3}\).

Step by step solution

01

In our function, \(f(x) = \ln(2x + 3)\), the outer function is \(\ln\) and the inner function is \(2x + 3\). #Step 2: Differentiate the outer function#

We have already identified the outer function as \(\ln\). If we represent the outer function as \(u = \ln(v)\), we have to find \(\frac{du}{dv}\). Recall that the derivative of the natural logarithm function is \(\frac{1}{v}\). Thus, we have \(\frac{du}{dv} = \frac{1}{v}\) #Step 3: Differentiate the inner function#
02

The inner function is \(2x + 3\) which we denote as \(v = 2x + 3\). Now we have to find \(\frac{dv}{dx}\). Using the basic rules of differentiation, we get \(\frac{dv}{dx} = 2\). #Step 4: Apply the chain rule#

Now we apply the chain rule \((f(g(x)))' = f'(g(x)) \cdot g'(x)\). In our case, this translates to \((f(v(x)))' = f'(v(x)) \cdot v'(x)\). We have found the derivatives \(f'(v) = \frac{1}{v}\) and \(v'(x) = 2\), so we need to substitute these values: \( f'(x) = \frac{1}{v} \cdot 2 \) #Step 5: Replace v with the inner function#
03

In Step 1, we determined that the inner function is \(v(x) = 2x+3\). Now let us replace \(v\) with this function: \( f'(x) = \frac{1}{2x+3} \cdot 2 \) #Step 6: Simplify the result#

We now simplify the result of our calculation: \( f'(x) = \frac{2}{2x+3} \) Finally, we have the derivative of the function \(f(x) = \ln(2x + 3)\): $$ f'(x) = \frac{2}{2x+3} $$

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Key Concepts

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

Natural Logarithm
The natural logarithm, denoted as \( \ln(x) \), is the logarithm that uses the base of the natural number \( e \), where \( e \approx 2.71828 \). This mathematical function is highly significant in calculus and various fields of science and engineering. The natural logarithm helps to transform multiplicative processes into additive processes, making them easier to analyze and solve.
When differentiating functions involving natural logarithms, it's important to remember that the derivative of \( \ln(u) \), where \( u \) is a function of \( x \), is \( \frac{1}{u} \cdot \frac{du}{dx} \). Essentially, this means you first differentiate the natural logarithm itself, which is \( \frac{1}{u} \), and then multiply by the derivative of the argument \( u \) with respect to \( x \).
Understanding natural logarithms and their properties is crucial when dealing with continuous growth processes and areas where base \( e \) exponential functions are prominent, such as in compound interest or population growth models.
Chain Rule
The chain rule is a fundamental technique in calculus used for finding the derivative of composite functions. A composite function is one where a function is applied inside another function, such as \( f(g(x)) \). The chain rule helps us differentiate these kinds of expressions by breaking them down into their inner and outer functions.
When you have a composite function, say \( y = u(v(x)) \), the chain rule states that the derivative \( \frac{dy}{dx} \) is the product of the derivative of the outer function \( u \) with respect to the inner function \( v \), \( \frac{du}{dv} \), and the derivative of the inner function \( v \) with respect to \( x \), \( \frac{dv}{dx} \). So, it is given by the formula:
  • \( \frac{dy}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} \)

In our exercise, \( f(x) = \ln(2x + 3) \), the chain rule becomes particularly handy. Here, the outer function is \( \ln(v) \) and the inner function is \( v = 2x + 3 \). The chain rule allowed us to first find \( \frac{1}{v} \) for the natural logarithm and multiply it with \( 2 \), the derivative of the inner function. This method of 'chaining' derivatives maintains the structured linkage of the function's parts.
Differentiation Rules
Differentiation rules are a set of guidelines that help us find the derivative of various types of functions quickly and accurately. These rules are especially useful when dealing with more complex functions like those seen in calculus involving polynomials, trigonometric functions, exponential functions, and logarithms.
Some commonly used differentiation rules include:
  • The Power Rule: \( \frac{d}{dx}[x^n] = nx^{n-1} \)
  • The Product Rule: \( \frac{d}{dx}[uv] = u'v + uv' \)
  • The Quotient Rule: \( \frac{d}{dx}\left[ \frac{u}{v} \right] = \frac{u'v - uv'}{v^2} \)
  • The Chain Rule, as discussed earlier

For the specific case of differentiating \( \ln(2x + 3) \), we apply the rules involved with logarithms and the chain rule. To handle the inner function, we use linearity in differentiation, where the derivative of \( 2x + 3 \) is simply \( 2 \), showing how basic differentiation rules simplify the process.
Understanding and using these rules effectively makes the calculus process more fluid and manageable, especially as functions increase in complexity.

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Most popular questions from this chapter

A projectile is fired from a cannon located on a horizontal plane. If we think of the cannon as being located at the origin \(O\) of an \(x y\) -coordinate system, then the path of the projectile is $$y=\sqrt{3} x-\frac{x^{2}}{400}$$ where \(x\) and \(y\) are measured in feet. a. Find the value of \(\theta\) (the angle of elevation of the gun). b. At what point on the trajectory is the projectile traveling parallel to the ground? c. What is the maximum height attained by the projectile? d. What is the range of the projectile (the distance \(O A\) along the \(x\) -axis)? e. At what angle with respect to the \(x\) -axis does the projectile hit the ground?

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In Exercises, (a) find the equations of the tangent and the normal lines to the curve at the indicated point. (The normal line at a point on the curve is the line perpendicular to the tangent line at that point.) (b) Then use a graphing utility to plot the curve and the tangent and normal lines on the same screen. $$ 4 x y-9=0 ; \quad\left(3, \frac{3}{4}\right) $$

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