Problem:
Differentiate the following function:
$\sqrt{-7 x-3}+\frac{1}{2 x+8}$
Answer:
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\begin{array}{l}
 \text{Possible derivation:} \\
 \frac{d}{dx}\left(\sqrt{-3-7 x}+\frac{1}{8+2 x}\right) \\
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 \\
\hline
 
\begin{array}{l}
 \text{Differentiate }\text{the }\text{sum }\text{term }\text{by }\text{term:} \\
  \text{= }\frac{d}{dx}\left(\sqrt{-3-7 x}\right)+\frac{d}{dx}\left(\frac{1}{8+2 x}\right) \\
\end{array}
 \\
 
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 \text{Using }\text{the }\text{chain }\text{rule, }\frac{d}{dx}\left(\sqrt{-7 x-3}\right)=\frac{d\sqrt{u}}{du} \frac{du}{dx}, \text{where }u=-7 x-3 \text{and }\frac{d\text{}}{du}\left(\sqrt{u}\right)=\frac{1}{2 \sqrt{u}}: \\
  \text{= }\frac{d}{dx}\left(\frac{1}{8+2 x}\right)+\fbox{$\frac{\frac{d}{dx}(-3-7 x)}{2 \sqrt{-3-7 x}}$} \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Differentiate }\text{the }\text{sum }\text{term }\text{by }\text{term }\text{and }\text{factor }\text{out }\text{constants:} \\
  \text{= }\frac{d}{dx}\left(\frac{1}{8+2 x}\right)+\fbox{$\frac{d}{dx}(-3)-7 \left(\frac{d}{dx}(x)\right)$} \frac{1}{2 \sqrt{-3-7 x}} \\
\end{array}
 \\
 
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 \text{The }\text{derivative }\text{of }-3 \text{is }\text{zero}: \\
  \text{= }\frac{d}{dx}\left(\frac{1}{8+2 x}\right)+\frac{-7 \left(\frac{d}{dx}(x)\right)+\fbox{$0$}}{2 \sqrt{-3-7 x}} \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Simplify }\text{the }\text{expression}: \\
  \text{= }-\frac{7 \left(\frac{d}{dx}(x)\right)}{2 \sqrt{-3-7 x}}+\frac{d}{dx}\left(\frac{1}{8+2 x}\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{The }\text{derivative }\text{of }x \text{is }1: \\
  \text{= }\frac{d}{dx}\left(\frac{1}{8+2 x}\right)-\fbox{$1$} \frac{7}{2 \sqrt{-3-7 x}} \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Using }\text{the }\text{chain }\text{rule, }\frac{d}{dx}\left(\frac{1}{2 x+8}\right)=\frac{d}{du}\frac{1}{u} \frac{du}{dx}, \text{where }u=2 x+8 \text{and }\frac{d\text{}}{du}\left(\frac{1}{u}\right)=-\frac{1}{u^2}: \\
  \text{= }-\frac{7}{2 \sqrt{-3-7 x}}+\fbox{$-\frac{\frac{d}{dx}(8+2 x)}{(8+2 x)^2}$} \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Differentiate }\text{the }\text{sum }\text{term }\text{by }\text{term }\text{and }\text{factor }\text{out }\text{constants:} \\
  \text{= }-\frac{7}{2 \sqrt{-3-7 x}}-\fbox{$\frac{d}{dx}(8)+2 \left(\frac{d}{dx}(x)\right)$} \frac{1}{(8+2 x)^2} \\
\end{array}
 \\
 
\begin{array}{l}
 \text{The }\text{derivative }\text{of }8 \text{is }\text{zero}: \\
  \text{= }-\frac{7}{2 \sqrt{-3-7 x}}-\frac{2 \left(\frac{d}{dx}(x)\right)+\fbox{$0$}}{(8+2 x)^2} \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Simplify }\text{the }\text{expression}: \\
  \text{= }-\frac{7}{2 \sqrt{-3-7 x}}-\frac{2 \left(\frac{d}{dx}(x)\right)}{(8+2 x)^2} \\
\end{array}
 \\
 
\begin{array}{l}
 \text{The }\text{derivative }\text{of }x \text{is }1: \\
  \text{= }-\frac{7}{2 \sqrt{-3-7 x}}-\fbox{$1$} \frac{2}{(8+2 x)^2} \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Simplify }\text{the }\text{expression}: \\
  \text{= }-\frac{7}{2 \sqrt{-3-7 x}}-\frac{2}{(8+2 x)^2} \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Simplify }\text{the }\text{expression}: \\
 \fbox{$
\begin{array}{ll}
 \text{Answer:} &  \\
 \text{} &  \text{= }-\frac{7}{2 \sqrt{-3-7 x}}-\frac{1}{2 (4+x)^2} \\
\end{array}
$} \\
\end{array}
 \\
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{'source': 'AMPS', 'file': 'amps/mathematica/calculus/derivatives_w_steps/923.txt'}