/*! 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 77 Use the properties of logarithms... [FREE SOLUTION] | 91Ó°ÊÓ

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Use the properties of logarithms to simplify the following functions before computing \(f^{\prime}(x)\). $$f(x)=\ln (3 x+1)^{4}$$

Short Answer

Expert verified
Answer: The simplified version of the function is \(f(x) = 4\,\ln(3x+1)\) and its derivative is \(f'(x) = \frac{12}{3x+1}\).

Step by step solution

01

Simplify the function using the properties of logarithms.

Use the property \(\ln(a^b) = b\, \ln(a)\): $$f(x) = \ln(3x+1)^4 = 4\,\ln(3x+1)$$ Now that we have a simplified expression for \(f(x)\), we can compute \(f'(x)\).
02

Compute the function's derivative

Use the chain rule to compute the derivative of the composite function: \((g(h(x)))' = g'(h(x))\,h'(x)\). Here, let \(g(x) = 4\,\ln(x)\) and \(h(x) = 3x+1\). Then, \(g'(x)\) and \(h'(x)\) are: $$g'(x) = \frac{4}{x} \quad\text{and}\quad h'(x) = 3$$ We can now apply the chain rule: $$f'(x) = 4\,\frac{1}{3x+1}\cdot 3 = \frac{12}{3x+1}$$ So, after using the properties of logarithms and applying the chain rule, the simplified function and its derivative are: $$f(x) = 4\,\ln(3x+1) \quad\text{and}\quad f'(x) = \frac{12}{3x+1}$$

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Key Concepts

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

Logarithmic Differentiation
Logarithmic differentiation is a powerful mathematical technique which simplifies the process of finding derivatives, especially when dealing with products, quotients, and powers of functions. It exploits the properties of logarithms to turn multiplication into addition, division into subtraction, and powers into products. This is particularly useful when dealing with complicated expressions that would be difficult to differentiate using basic rules.

In the given exercise, the function presented is \( f(x) = \ln(3x+1)^4 \). Using the property of logarithms that allows us to bring down the exponent (\(\ln(a^b) = b \, \ln(a)\)), we simplify the expression to \(4\ln(3x+1)\), which is easier to handle when taking the derivative. This initial step is the essence of logarithmic differentiation: simplify first, calculate the derivative second.

Once the expression is simplified, we move on to apply the differentiation rules, confident that the mathematical 'heavy lifting' has been managed through our clever use of logarithmic properties. This makes calculating derivatives more manageable and streamlined, and it's a testament to logarithmic differentiation's utility in calculus.
Chain Rule in Calculus
The chain rule is among the most important rules in calculus for finding derivatives of composite functions. When we have a function composed of one function inside another, we can't differentiate each part separately. Instead, we use the chain rule, which provides a method to differentiate the entire composite function systematically.

The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. Mathematically, this is represented as \( (g(h(x)))' = g'(h(x)) \, h'(x) \).

In our exercise, \( g(x) = 4\ln(x) \) and \( h(x) = 3x+1 \). By finding \( g'(x) = \frac{4}{x} \) and \( h'(x) = 3 \) first, we can then use the chain rule to effectively find \( f'(x) \) as \(4 \frac{1}{3x+1} \cdot 3\), or simplified to \( \frac{12}{3x+1} \).

Utilizing the chain rule in this manner allows students to tackle more complex derivatives methodically, breaking down the problem into manageable steps and ultimately revealing the simplicity within the complexity.
Derivative of Logarithmic Functions
Logarithmic functions pose their own unique challenges in calculus, but once their derivative principles are understood, they can be approached with confidence. The derivative of the natural logarithm function, \(\ln(x)\), is one of the fundamental derivatives in calculus and is equal to \( \frac{1}{x} \).

However, when working with logarithmic functions that include additional components, such as coefficients or composite functions, the differentiation process becomes slightly more involved. For example, if the function is a multiplied constant times the natural logarithm, as in our example \( f(x) = 4\ln(3x+1) \), the constant can be treated separately from the \(\ln \) function.

To find \(f'(x)\) of our function, we apply the derivative of the natural logarithm to the inner function \(3x+1\), multiplying it by the derivative of the inner function itself, which is 3. The constant 4 simply remains as a constant factor in the derivative, giving us the final result of \( \frac{12}{3x+1} \). This consistent approach to logarithmic differentiation fluently combines the understanding of logarithms with the mechanics of differentiation to effectively deal with a variety of functions.

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Most popular questions from this chapter

\(F=f / g\) be the quotient of two functions that are differentiable at \(x\) a. Use the definition of \(F^{\prime}\) to show that $$ \frac{d}{d x}\left(\frac{f(x)}{g(x)}\right)=\lim _{h \rightarrow 0} \frac{f(x+h) g(x)-f(x) g(x+h)}{h g(x+h) g(x)} $$ b. Now add \(-f(x) g(x)+f(x) g(x)\) (which equals 0 ) to the numerator in the preceding limit to obtain $$\lim _{h \rightarrow 0} \frac{f(x+h) g(x)-f(x) g(x)+f(x) g(x)-f(x) g(x+h)}{h g(x+h) g(x)}$$ Use this limit to obtain the Quotient Rule. c. Explain why \(F^{\prime}=(f / g)^{\prime}\) exists, whenever \(g(x) \neq 0\)

Work carefully Proceed with caution when using implicit differentiation to find points at which a curve has a specified slope. For the following curves, find the points on the curve (if they exist) at which the tangent line is horizontal or vertical. Once you have found possible points, make sure they actually lie on the curve. Confirm your results with a graph. $$x^{2}\left(3 y^{2}-2 y^{3}\right)=4$$

a. Use derivatives to show that \(\tan ^{-1} \frac{2}{n^{2}}\) and \(\tan ^{-1}(n+1)-\tan ^{-1}(n-1)\) differ by a constant. b. Prove that \(\tan ^{-1} \frac{2}{n^{2}}=\tan ^{-1}(n+1)-\tan ^{-1}(n-1)\)

Suppose \(f\) is differentiable on an interval containing \(a\) and \(b\), and let \(P(a, f(a))\) and \(Q(b, f(b))\) be distinct points on the graph of \(f\). Let \(c\) be the \(x\) -coordinate of the point at which the lines tangent to the curve at \(P\) and \(Q\) intersect, assuming that the tangent lines are not parallel (see figure). a. If \(f(x)=x^{2},\) show that \(c=(a+b) / 2,\) the arithmetic mean of \(a\) and \(b\), for real numbers \(a\) and \(b\) b. If \(f(x)=\sqrt{x}\), show that \(c=\sqrt{a b}\), the geometric mean of \(a\) and \(b\), for \(a > 0\) and \(b > 0\) c. If \(f(x)=1 / x,\) show that \(c=2 a b /(a+b),\) the harmonic mean of \(a\) and \(b,\) for \(a > 0\) and \(b > 0\) d. Find an expression for \(c\) in terms of \(a\) and \(b\) for any (differentiable) function \(f\) whenever \(c\) exists.

General logarithmic and exponential derivatives Compute the following derivatives. Use logarithmic differentiation where appropriate. $$\frac{d}{d x}\left(x^{\left(x^{10}\right)}\right)$$

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