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Find the following derivatives. $$\frac{d}{d x}(\ln (\ln x))$$

Short Answer

Expert verified
Answer: The derivative of the composite function \(\ln(\ln x)\) with respect to \(x\) is \(\frac{1}{x \ln x}\).

Step by step solution

01

Find the derivative of the outer function

The outer function is \(u(x) = \ln x\). We know that the derivative of this function with respect to \(x\) is: $$u'(x) = \frac{1}{x}$$
02

Find the derivative of the inner function

The inner function is \(v(x) = \ln x\). As we computed in step 1, the derivative of this function with respect to \(x\) is: $$v'(x) = \frac{1}{x}$$
03

Apply the chain rule

According to the chain rule, the derivative of the composite function \(\ln (\ln x)\) with respect to \(x\) is: $$\frac{d}{d x}(\ln (\ln x)) = u'(v(x)) \cdot v'(x)$$ Plugging in our expressions for \(u'(x)\), \(v(x)\), and \(v'(x)\), we get: $$\frac{d}{d x}(\ln (\ln x)) = \left(\frac{1}{\ln x}\right) \cdot \frac{1}{x}$$ So, the derivative is: $$\frac{d}{d x}(\ln (\ln x)) = \frac{1}{x \ln x}$$

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Key Concepts

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

Chain Rule
Understanding the chain rule is crucial for calculating derivatives of composite functions. In calculus, many functions are composed of other functions, which means they can be broken down into an outside function and an inside function. The chain rule tells us how to take the derivative of such composite functions. In essence, it states to differentiate the outside function first, evaluate it at the inside function, and then multiply by the derivative of the inside function.

When we apply the chain rule to the function \( \ln(\ln x) \), we are dealing with two nested natural logarithms. Here, we consider the outer function to be \( u(x) = \ln x \) and the inner function to be \( v(x) = \ln x \). The chain rule simplifies what would otherwise be a complicated derivative into a manageable process, multiplying the derivative of the outer function by the derivative of the inner function, which lets us find that \( \frac{d}{dx}(\ln(\ln x)) = \frac{1}{x \ln x} \).
Natural Logarithm
The natural logarithm, denoted as \( \ln x \), is a fundamental concept in calculus and represents the logarithm to the base \( e \), where \( e \approx 2.718 \). It is the inverse function of the exponential function \( e^x \). In terms of derivatives, the natural logarithm is noteworthy because its derivative is relatively simple: \( \frac{d}{dx}\ln x = \frac{1}{x} \), given \( x > 0 \).

When encountering problems such as \( \frac{d}{dx}(\ln(\ln x)) \), recognizing that the derivative of a \( \ln \) function follows this straightforward pattern allows for the application of the chain rule more confidently. This also illustrates why natural logarithms are so prevalent in calculus—they make many complex derivative problems more tractable.
Derivatives in Calculus
Derivatives are a core component in calculus and provide a method to determine the instantaneous rate of change of a function with respect to a variable—essentially, they tell us how a function is changing at any given point. The derivative is found using a variety of rules and formulas, depending on the type of function involved.

For instance, the derivative of a polynomial function involves applying the power rule, while trigonometric functions have their own specific derivatives. Differentiation becomes even more interesting when involving composite functions, where strategies like the chain rule, the product rule, and the quotient rule come into play. As seen in the exercise involving \( \ln(\ln x) \), the derivative tells us how the expression \( \ln(\ln x) \) changes as \( x \) changes, highlighting the interconnectedness of the function and its rate of change.

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Most popular questions from this chapter

Find the following higher-order derivatives. $$\left.\frac{d^{3}}{d x^{3}}\left(x^{4.2}\right)\right|_{x=1}$$

Find the following higher-order derivatives. $$\frac{d^{3}}{d x^{3}}\left(x^{2} \ln x\right)$$.

The total energy in megawatt-hr (MWh) used by a town is given by $$E(t)=400 t+\frac{2400}{\pi} \sin \frac{\pi t}{12},$$ where \(t \geq 0\) is measured in hours, with \(t=0\) corresponding to noon. a. Find the power, or rate of energy consumption, \(P(t)=E^{\prime}(t)\) in units of megawatts (MW). b. At what time of day is the rate of energy consumption a maximum? What is the power at that time of day? c. At what time of day is the rate of energy consumption a minimum? What is the power at that time of day? d. Sketch a graph of the power function reflecting the times at which energy use is a minimum or maximum.

Let \(f\) and \(g\) be differentiable functions with \(h(x)=f(g(x)) .\) For a given constant \(a,\) let \(u=g(a)\) and \(v=g(x),\) and define $$H(v)=\left\\{\begin{array}{ll} \frac{f(v)-f(u)}{v-u}-f^{\prime}(u) & \text { if } v \neq u \\ 0 & \text { if } v=u. \end{array}\right.$$ a. Show that \(\lim _{x \rightarrow u} H(v)=0\) b. For any value of \(u\) show that $$f(v)-f(u)=\left(H(v)+f^{\prime}(u)\right)(v-u).$$ c. Show that. $$h^{\prime}(a)=\lim _{x \rightarrow a}\left(\left(H(g(x))+f^{\prime}(g(a))\right) \cdot \frac{g(x)-g(a)}{x-a}\right).$$ d. Show that \(h^{\prime}(a)=f^{\prime}(g(a)) g^{\prime}(a)\).

Vertical tangent lines a. Determine the points at which the curve \(x+y^{3}-y=1\) has a vertical tangent line (see Exercise 52 ). b. Does the curve have any horizontal tangent lines? Explain.

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