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 $f'$.
- $\displaystyle f(x)=x^2+\frac{1}{x^2}-\frac{5}{\sqrt[3]{x}}$
- $\displaystyle f(x)=x^2e^{\sqrt{x}}$
- $\displaystyle f(x)=x^{\cos(x)}$
- $\displaystyle f(x)=\frac{e^x-1}{e^x+1}$
- $\displaystyle f(x)=\sqrt{\log_{10}(x+4)}$
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.
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 $\mathbb{D}_{f}=\mathbb{D}_{f'}=\mathbb{R}^+$.
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 $\mathbb{D}_{f}=\mathbb{D}_{f'}=\mathbb{R}^+_0$
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.
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.
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. $\begin{align*}f'(x) &= \frac{\left(\sqrt{\ln(x+4)}\right)'}{\sqrt{\ln(10)}}\\ &=\frac{\frac{1}{2\sqrt{\ln(x+4)}}\cdot\frac{1}{(x+4)}}{\sqrt{\ln(10)}}\\ &=\frac{\sqrt{\ln(10)}}{(2x+8)\sqrt{\ln(x+4)}}\\ &= \frac{1}{(2x+8)\sqrt{\ln(x+4)}\sqrt{\ln(10)}}.\end{align*}$