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Show that \(\frac{d}{d x}(\ln k x)=\frac{d}{d x}(\ln x),\) where \(x>0\) and \(k\) is a positive real number.

Short Answer

Expert verified
Question: Show that \(\frac{d}{d x}(\ln k x)=\frac{d}{d x}(\ln x),\) where \(x>0\) and \(k\) is a positive real number. Answer: \(\frac{d}{d x}(\ln k x) = \frac{1}{x}\) and \(\frac{d}{d x}(\ln x) = \frac{1}{x}\), so the derivatives are equal. Therefore, \(\frac{d}{d x}(\ln k x)=\frac{d}{d x}(\ln x)\).

Step by step solution

01

Differentiate \(\ln(kx)\) with respect to \(x\)

To differentiate \(\ln(kx)\) with respect to \(x\), use the chain rule. The chain rule states that \((f(g(x)))' = f'(g(x)) \cdot g'(x)\). Let \(f(u) = \ln(u)\) and \(g(x) = kx\). Then, we have: \(f'(u) = \frac{1}{u}\) (this is the derivative of the natural logarithm) \(g'(x) = k\) (this is the derivative of a constant times a variable) Now, apply the chain rule: \((f(g(x)))' = f'(g(x)) \cdot g'(x) = \frac{1}{kx} \cdot k\) Simplify: \(\frac{d}{d x}(\ln k x) = \frac{k}{kx} = \frac{1}{x}\)
02

Differentiate \(\ln(x)\) with respect to \(x\)

The derivative of \(\ln(x)\) with respect to \(x\) is: \(\frac{d}{d x}(\ln x) = \frac{1}{x}\)
03

Compare the derivatives

Now compare the two derivatives. We have: \(\frac{d}{d x}(\ln k x) = \frac{1}{x}\) and \(\frac{d}{d x}(\ln x) = \frac{1}{x}\) Since the derivatives are equal, we have shown the required result: \(\frac{d}{d x}(\ln k x) = \frac{d}{d x}(\ln x)\)

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Key Concepts

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

Chain Rule Calculus
Understanding the chain rule in calculus is vital for differentiating composite functions, which are functions composed of two or more functions. The chain rule can be stated as follows: If you have a function \(f(g(x))\), where \(f\) and \(g\) are both functions of \(x\), then the derivative of this composite function with respect to \(x\) is \(f'(g(x)) \times g'(x)\).

When differentiating a function like \(\ln(kx)\), where \(k\) is a constant, the chain rule allows us to differentiate the inner function \(g(x) = kx\) and the outer function \(f(u) = \ln(u)\) separately, and then multiply their derivatives to get the derivative of the whole composite function.

In simpler terms, we first differentiate \(\ln(u)\) concerning \(u\), which gives us \(\frac{1}{u}\), and then differentiate \(kx\) concerning \(x\), which gives us \(k\). These two results are then multiplied to give the derivative of \(\ln(kx)\) with respect to \(x\).
Natural Logarithm Derivative
The natural logarithm function, commonly denoted as \(\ln(x)\), is one whose derivative is crucial in calculus. The derived formula for the natural logarithm is quite simple: for any positive number \(x\), the derivative of \(\ln(x)\) with respect to \(x\) is \(\frac{1}{x}\).

This rule stands as one of the fundamental building blocks when dealing with logarithmic differentiation. When you understand that the derivative of \(\ln(x)\) is \(\frac{1}{x}\), you can easily apply this knowledge to more complex logarithmic functions by using the chain rule for differentiation, as seen in the exercise above.
Implicit Differentiation
Implicit differentiation is a technique that is often applied when dealing with equations where the dependent variable, typically \(y\), cannot be easily isolated. In such cases, you assume that \(y\) is a function of \(x\) and differentiate both sides of the equation with respect to \(x\), applying the chain rule where necessary.

Even if this concept was not directly applied in the original exercise, it is important to understand, as implicitly differentiating functions is another method of finding the derivative without explicitly solving for \(y\). In the context of logarithmic differentiation, implicit differentiation can be particularly handy when dealing with more complex equations involving logarithms.
Calculus Proofs
Calculus proofs are rigorous arguments that confirm the validity of mathematical statements within calculus. Proofs may involve demonstrating that a certain differentiation strategy works, as in the textbook exercise example.

In the given exercise, the proof involved showing that \(\frac{d}{dx}(\ln(kx)) = \frac{d}{dx}(\ln(x))\), regardless of the value of constant \(k\). The proof required knowledge of derivative rules, including the chain rule, and an understanding of the natural logarithm derivative. By breaking down the proof into comprehensible steps, demonstrating each differentiation process, the result verifies the proposition with clarity and helps students understand the underpinning concepts more fully.

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Most popular questions from this chapter

Logistic growth Scientists often use the logistic growth function \(P(t)=\frac{P_{0} K}{P_{0}+\left(K-P_{0}\right) e^{-r_{d}}}\) to model population growth, where \(P_{0}\) is the initial population at time \(t=0, K\) is the carrying capacity, and \(r_{0}\) is the base growth rate. The carrying capacity is a theoretical upper bound on the total population that the surrounding environment can support. The figure shows the sigmoid (S-shaped) curve associated with a typical logistic model. World population (part 1 ) The population of the world reached 6 billion in \(1999(t=0)\). Assume Earth's carrying capacity is 15 billion and the base growth rate is \(r_{0}=0.025\) per year. a. Write a logistic growth function for the world's population (in billions) and graph your equation on the interval \(0 \leq t \leq 200\) using a graphing utility. b. What will the population be in the year 2020? When will it reach 12 billion?

Recall that \(f\) is even if \(f(-x)=f(x),\) for all \(x\) in the domain of \(f,\) and \(f\) is odd if \(f(-x)=-f(x),\) for all \(x\) in the domain of \(f\). a. If \(f\) is a differentiable, even function on its domain, determine whether \(f^{\prime}\) is even, odd, or neither. b. If \(f\) is a differentiable, odd function on its domain, determine whether \(f^{\prime}\) is even, odd, or neither.

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.

The lateral surface area of a cone of radius \(r\) and height \(h\) (the surface area excluding the base) is \(A=\pi r \sqrt{r^{2}+h^{2}}\) a. Find \(d r / d h\) for a cone with a lateral surface area of $$ A=1500 \pi $$ b. Evaluate this derivative when \(r=30\) and \(h=40\)

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. $$x^{2}(y-2)-e^{y}=0$$

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