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Derive the following functions and simplify as good as possible. Then determine the maximum domain (over $\mathbb{R}$) of each function $f$ and its derivative $f'$.

  1. $\displaystyle f(x)=x^2+\frac{1}{x^2}-\frac{5}{\sqrt[3]{x}}$
  2. $\displaystyle f(x)=x^2e^{\sqrt{x}}$
  3. $\displaystyle f(x)=x^{\cos(x)}$
  4. $\displaystyle f(x)=\frac{e^x-1}{e^x+1}$
  5. $\displaystyle f(x)=\sqrt{\log_{10}(x+4)}$

I think I have all derivatives correct and therefore I will shorten things a bit. What I would like to know is whether I can simplify my results I have by now. Furthermore I have some domains (denoted by $\mathbb{D}_f$ and $\mathbb{D}_{f'}$) by guessing them, however I have some issues with a few of them.

  1. With $(x^{-1/3})'=-1/3\cdot x^{-4/3}=-1/3x\sqrt[3]{x}$ it's fairly simple to see that $$f'(x) = 2x-\frac{2}{x^3}+\frac{5}{3x\sqrt[3]{x}}.$$ Therefore $\mathbb{D}_{f}=\mathbb{D}_{f'}=\mathbb{R}^+$.

  2. Using the product rule and by multiplying with 2 and 1/2 respectively we can then simplify the result $$\begin{align*}f'(x) &= 2xe^{\sqrt{x}}+x^2\frac{1}{2\sqrt{x}}e^{\sqrt{x}}\\&= xe^{\sqrt{x}}\left(2+\frac{x}{2\sqrt{x}}\right)\\ &=\frac{1}{2}xe^{\sqrt{x}}(4+\sqrt{x}).\end{align*}$$ Based on the square root it is obvious that $\mathbb{D}_{f}=\mathbb{D}_{f'}=\mathbb{R}^+_0$

  3. Recognizing $f(x)=x^{\cos(x)}=e^{\ln(x^{\cos(x)})}=e^{\cos(x)\ln(x)}$ we can use the chain rule and the product rule to derive the function. $$\begin{align*}f'(x)&=(\cos(x)\ln(x))'x^{\cos(x)}\\&=\left(-\sin(x)\ln(x)+\cos(x)\frac{1}{x}\right)x^{\cos(x)}\\&=(-x\ln(x)\sin(x)+\cos(x))x^{\cos(x)-1}.\end{align*}$$ To prevent issues with some square roots and negative number i think that $\mathbb{D}_{f}=\mathbb{D}_{f'}=\mathbb{R}^+_0$ but I am not sure at all.

  4. By the quotient rule we get $$\begin{align*}f'(x) &=\frac{e^x(e^x+1)-(e^x-1)e^x}{(e^x+1)^2}\\ &=\frac{2e^x}{(e^x+1)^2}\\ &=\frac{1}{\cosh(x)+1}.\end{align*}$$ I don't know how to exactly determine the domains because I think that the $\exp$ function is defined for $\mathbb{R}$ yet this seems to easy to be true.

  5. Using the chain rule and the derivative for the $\log$ we can easily compute the derivative though I have no idea aboout the restrictions of the domain. $$\begin{align*}f'(x) &= \frac{\left(\sqrt{\ln(x+4)}\right)'}{\sqrt{\ln(10)}}\\ &=\frac{\frac{1}{2\sqrt{\ln(x+4)}}\cdot\frac{1}{(x+4)}}{\sqrt{\ln(10)}}\\ &=\frac{\sqrt{\ln(10)}}{(2x+8)\sqrt{\ln(x+4)}}\\ &= \frac{1}{(2x+8)\sqrt{\ln(x+4)}\sqrt{\ln(10)}}.\end{align*}$$

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I do not agree with you on 1. Everybody agrees that $\sqrt[3]{x}$ is defined for every $x \in \mathbb{R}$. –  Siminore Dec 3 '12 at 13:00
@Siminore: Yet one can argument that e.g. $\sqrt{\sqrt{-8}}$ leads to problems if we don't care about complex numbers (which are excluded in this excercise - I did forgot to mention this), or am I wrong here? –  Christian Ivicevic Dec 3 '12 at 13:21
$\sqrt{x}$ and $\sqrt[3]{x}$ are rather different things, you know. For example $\sqrt{-1} \notin \mathbb{R}$, while $\sqrt[3]{-1}=-1$. –  Siminore Dec 3 '12 at 14:02
@Siminore: Do you have some other clues about the domains of (3)-(5)? –  Christian Ivicevic Dec 3 '12 at 14:15
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1 Answer

up vote 0 down vote accepted

I was now able to find all domains.

  1. $\mathbb{D}_f=\mathbb{D}_{f'}=\mathbb{R}\setminus \{0\}$ - the cube root is defined for all reals however we need to pay attention that our fraction do not get $1/0$.
  2. $\mathbb{D}_f=\mathbb{D}_{f'}=\mathbb{R}^+_0$ - based on the square root.
  3. $\mathbb{D}_f=\mathbb{D}_{f'}=\mathbb{R}^+$ - based on the $\ln$ in both functions which only accepts parameters $x>0$.
  4. $\mathbb{D}_f=\mathbb{D}_{f'}=\mathbb{R}$ - we know that $\exp$ is defined for all reals and always positive. Therefore we cannot run into any problem to get a denominator which is 0. The same applies for the derivative.
  5. $\mathbb{D}_f=\mathbb{D}_{f'}=[-3, \infty)$ - the only thing implying this is the $\sqrt{\ln(x+4)}$ term. We know that $\ln(x+4)\ge 0\implies x\ge -3$ and we have our domains.

I hope this suffices for now. Any suggestions are appreciated!

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In 3, the condition $x>0$ is a convention to ease our lives. Strictly speaking, there may be negative numbers at which $x^{\cos x}$ can be computed in $\mathbb{R}$, but nobody cares about them. –  Siminore Dec 3 '12 at 15:18
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