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The operation of Problem $4.2.3$ is sometimes viewed as a $"$derivative and $a/L$ is written $\frac{dL}{da}.$ These derivatives apply to regular expressions in a manner similar to the way ordinary derivatives apply to arithmetic expressions. Thus if $R$ is a regular expression, we shall use $\frac{dR}{da}$ to mean the same as $\frac{dL}{da},$if $L=L(R).$

- Show that $\frac{d(R+S)}{da}=\frac{dR}{da}+\frac{dS}{da}$
- Give the rule for the $"$derivative$"$ of $RS.$ Hint$:$ You need to consider two cases$:$ if $L(R)$ does or does not contain $\in.$ This rule is not quite the same as the $"$product rule$"$ for ordinary derivatives, but is similar.
- Give the rule for the $"$derivative$"$ of a closure, i.e.,$\frac{d(R^{*})}{da}$
- Use the rules from $(a)-(c)$ to find the $"$derivatives of regular expression $(0+1)^{*}011$ with respect to $$0 and $1.$
- Characterize those languages $L$ for which $\frac{dL}{d0}=\phi$
- Characterize those languages $L$ for which $\frac{dL}{d0}=L$

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