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Chapter 10: Problem 6

Berechnen Sie die Ableitungen der Funktionen \(f: D \rightarrow \mathbb{R}\) mit $$ \begin{aligned} &f_{1}(x)=\left(x+\frac{1}{x}\right)^{2}, \quad x \neq 0 \\ &f_{2}(x)=\cos \left(x^{2}\right) \cos ^{2} x, \quad x \in \mathbb{R} \\ &f_{3}(x)=\ln \left(\frac{\mathrm{e}^{x}-1}{\mathrm{e}^{x}}\right), \quad x \neq 0 \\ &f_{4}(x)=x^{\left(x^{x}\right)}, \quad x>0 \end{aligned} $$ auf dem jeweiligen Definitionsbereich der Funktion.

Short Answer

Expert verified
Given the 4 functions and their respective derivatives: 1. Function \(f_1(x) = (x + \frac{1}{x})^2\): The derivative is \(f_1'(x) = \frac{2(x^2 - 1)(x^2 + 1)}{x^3}\), with the domain \(x \neq 0\). 2. Function \(f_2(x) = \cos(x^2)\cos^2(x)\): The derivative is \(f_2'(x) = -4x \sin x^2 \cos^2 x - 2 \cos x^2 \cos x \sin x\), with the domain \(x \in \mathbb{R}\). 3. Function \(f_3(x) = \ln(\frac{e^x - 1}{e^x})\): The derivative is \(f_3'(x) = \frac{-1}{e^x(e^x - 1)}\), with the domain \(x \neq 0\). 4. Function \(f_4(x) = x^{x^x}, x > 0\): The derivative is \(f_4'(x)=x^{x^x}(x^{x^x}\left(\ln x + \frac{1}{x}\right))\), with the domain \(x > 0\). Find the derivatives of these functions in their respective domains of definition.

Step by step solution

01

Function 1: f1(x) = (x + 1/x)^2

Using the chain rule for differentiation, we have: \(f_1'(x) = 2(x + \frac{1}{x}) \cdot (1 - \frac{1}{x^2})\). Now, we can simplify the expression for \(f_1'(x)\): \(f_1'(x) = 2\left(x + \frac{1}{x}\right)\left(\frac{x^2 - 1}{x^2}\right) = 2\left(\frac{x^2 + 1}{x}\right)\left(\frac{x^2 - 1}{x^2}\right)\). Therefore, the derivative of \(f_1(x)\) is: $$f_1'(x) = \frac{2(x^2 - 1)(x^2 + 1)}{x^3}, \quad x \neq 0$$.
02

Function 2: f2(x) = cos(x^2)cos^2(x)

To find the derivative of \(f_2(x)\), we must use the product rule and chain rule: \(f_2'(x) = (\cos x^2)' \cdot \cos^2x + \cos x^2 \cdot (\cos^2 x)'\). Now, we can compute the derivatives of the individual functions and simplify: \(f_2'(x) = (-\sin x^2)(2x) \cdot \cos^2 x + \cos x^2 \cdot (2 \cos x)(-\sin x)\). Therefore, the derivative of \(f_2(x)\) is: $$f_2'(x) = -4x \sin x^2 \cos^2 x - 2 \cos x^2 \cos x \sin x, \quad x \in \mathbb{R}$$.
03

Function 3: f3(x) = ln((e^x - 1) / e^x)

To find the derivative of \(f_3(x)\), we must use the chain rule and the quotient rule: \(f_3'(x) = \frac{1}{\frac{e^x - 1}{e^x}} \cdot \left(\frac{e^x - e^x}{e^{2x}}\right)\). Simplifying, we get: \(f_3'(x) = \frac{e^x}{e^x - 1} \cdot \frac{-1}{e^x}\). Therefore, the derivative of \(f_3(x)\) is: $$f_3'(x) = \frac{-1}{e^x(e^x - 1)}, \quad x \neq 0$$.
04

Function 4: f4(x) = x^(x^x), x > 0

To find the derivative of \(f_4(x)\), we must first rewrite the function as \(f_4(x)= e^{\ln(x^{x^x})}\) and then use the chain rule: \(f_4'(x)= e^{\ln(x^{x^x})}\cdot\frac{d}{dx}\ln(x^{x^x})\). For the derivative of the natural logarithm, we use the chain rule again: \(\frac{d}{dx}\ln(x^{x^x}) = \frac{x^{x^x}}{x^{x^x}}\cdot\frac{d}{dx}(x^{x^x})\). Now, we need to compute the derivative of the exponent x^(x^x) \(\frac{d}{dx}(x^{x^x}) = (x^{x^x})\left(\frac{d}{dx}(x^x\ln x)\right)\). To compute the last derivative, we can use the product rule: \(\frac{d}{dx}(x^x\ln x) = (x^x)'(\ln x) + x^x(\frac{1}{x})\). We then simplify the expression: \((x^x)'=x^x\ln x + x^x\). Finally, we put all the terms back together to find the derivative of \(f_4(x)\): \(f_4'(x)=x^{x^x}\left(x^{x^x}\left(\ln x + \frac{1}{x}\right)\right)\). Therefore, the derivative of \(f_4(x)\) is: $$f_4'(x)=x^{x^x}(x^{x^x}\left(\ln x + \frac{1}{x}\right)),\quad x>0$$.

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Most popular questions from this chapter

Zeigen Sie f眉r \(|x|<1\) die Taylorformel$$ \ln \frac{1-x}{1+x}=-2\left(x+\frac{x^{3}}{3}+\cdots+\frac{x^{2 n-1}}{2 n-1}\right)+R_{2 n}(x) $$ mit dem Restglied $$ R_{2 n}(x)=\frac{-x^{2 n+1}}{2 n+1}\left(\frac{1}{(1+t x)^{2 n+1}}+\frac{1}{(1-t x)^{2 n+1}}\right) $$ f眉r ein \(t \in(0,1)\). Approximieren Sie mithilfe des Taylorpolynoms vom Grad \(n=\) 2 den Wert \(\ln (2 / 3)\) und zeigen Sie, dass der Fehler kleiner als \(5 \cdot 10^{-4}\) ist.

bullet Zeigen Sie durch eine vollst盲ndige Induktion die Ableitungen $$ \frac{\mathrm{d}^{n}}{\mathrm{~d} x^{n}}\left(\mathrm{e}^{x} \sin x\right)=(\sqrt{2})^{n} \mathrm{e}^{x} \sin \left(x+\frac{n \pi}{4}\right) $$ f眉r \(n=0,1,2, \ldots\)

Es wird eine H盲ngebr眉cke 眉ber eine \(30 \mathrm{~m}\) breiteBucht gebaut. Dabei ist die Form der Br眉cke durch die sogenannte Kettenlinie beschrieben, d.h., es gibt eine positive Konstante \(a>0\), sodass die Form durch den Graphen der Funktion \(f: \mathbb{R} \rightarrow \mathbb{R}\) mit $$ f(x)=h_{0}+a\left(\cosh \left(\frac{x-x_{0}}{a}\right)-1\right) $$ gegeben ist. Die Durchfahrtsh枚he f眉r Segelschiffe muss \(h_{0}=\) \(8 \mathrm{~m}\) betragen. Die Steilufer sind \(10 \mathrm{~m}\) und \(12 \mathrm{~m}\) 眉ber dem Wasserspiegel (siehe Abbildung). Bestimmen Sie mithilfe des Newton-Verfahrens den Parameter \(a>0\) und den Abstand \(x_{0}\) des Tiefpunkts zu einem der Ufer.

Berechnen Sie die Grenzwerte $$ \begin{aligned} &\lim _{x \rightarrow \infty} \frac{\ln (\ln x)}{\ln x} \\ &\lim _{x \rightarrow a} \frac{x^{a}-a^{x}}{a^{x}-a^{a}}, \quad \text { mit } a \in \mathbb{R}_{>0} \backslash\\{1\\} \\ &\lim _{x \rightarrow 0} \frac{1}{e^{r}-1}-\frac{1}{x} \\ &\lim _{x \rightarrow 0} \cot (x)(\arcsin (x)) \end{aligned} $$

Bestimmen Sie die Taylorreihe zu \(f: \mathbb{R} \rightarrow \mathbb{R}\) mit\(f(x)=x \exp (x-1)\) um \(x=1\) zum einen direkt und andererseits mithilfe der Potenzreihe zur Exponentialfunktion. Untersuchen Sie weiterhin die Reihe auf Konvergenz.
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