Problem:
Differentiate the following function:
$-e^{3 x+\frac{3}{2}} \csc (1-8 x)$
Answer:
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 \text{Possible derivation:} \\
 \frac{d}{dx}\left(-\left(e^{3/2+3 x} \csc (1-8 x)\right)\right) \\
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 \\
\hline
 
\begin{array}{l}
 \text{Factor }\text{out }\text{constants:} \\
  \text{= }-\left(\frac{d}{dx}\left(e^{3/2+3 x} \csc (1-8 x)\right)\right) \\
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 \\
 
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 \text{Use }\text{the }\text{product }\text{rule, }\frac{d}{dx}(u v)=v \frac{du}{dx}+u \frac{dv}{dx}, \text{where }u=e^{3 x+3/2} \text{and }v=\csc (1-8 x): \\
  \text{= }-\fbox{$\csc (1-8 x) \left(\frac{d}{dx}\left(e^{3/2+3 x}\right)\right)+e^{3/2+3 x} \left(\frac{d}{dx}(\csc (1-8 x))\right)$} \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Simplify }\text{the }\text{expression}: \\
  \text{= }-\csc (1-8 x) \left(\frac{d}{dx}\left(e^{3/2+3 x}\right)\right)-e^{3/2+3 x} \left(\frac{d}{dx}(\csc (1-8 x))\right) \\
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 \\
 
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 \text{Using }\text{the }\text{chain }\text{rule, }\frac{d}{dx}\left(e^{3 x+3/2}\right)=\frac{de^u}{du} \frac{du}{dx}, \text{where }u=3 x+\frac{3}{2} \text{and }\frac{d\text{}}{du}\left(e^u\right)=e^u: \\
  \text{= }-e^{3/2+3 x} \left(\frac{d}{dx}(\csc (1-8 x))\right)-\fbox{$e^{3/2+3 x} \left(\frac{d}{dx}\left(\frac{3}{2}+3 x\right)\right)$} \csc (1-8 x) \\
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 \\
 
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 \text{Differentiate }\text{the }\text{sum }\text{term }\text{by }\text{term }\text{and }\text{factor }\text{out }\text{constants:} \\
  \text{= }-e^{3/2+3 x} \left(\frac{d}{dx}(\csc (1-8 x))\right)-\fbox{$\frac{d}{dx}\left(\frac{3}{2}\right)+3 \left(\frac{d}{dx}(x)\right)$} e^{3/2+3 x} \csc (1-8 x) \\
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 \\
 
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 \text{The }\text{derivative }\text{of }\frac{3}{2} \text{is }\text{zero}: \\
  \text{= }-e^{3/2+3 x} \left(\frac{d}{dx}(\csc (1-8 x))\right)-e^{3/2+3 x} \csc (1-8 x) \left(3 \left(\frac{d}{dx}(x)\right)+\fbox{$0$}\right) \\
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 \\
 
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 \text{Simplify }\text{the }\text{expression}: \\
  \text{= }-3 e^{3/2+3 x} \csc (1-8 x) \left(\frac{d}{dx}(x)\right)-e^{3/2+3 x} \left(\frac{d}{dx}(\csc (1-8 x))\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{The }\text{derivative }\text{of }x \text{is }1: \\
  \text{= }-e^{3/2+3 x} \left(\frac{d}{dx}(\csc (1-8 x))\right)-\fbox{$1$} 3 e^{3/2+3 x} \csc (1-8 x) \\
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 \\
 
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 \text{Using }\text{the }\text{chain }\text{rule, }\frac{d}{dx}(\csc (1-8 x))=\frac{d\csc (u)}{du} \frac{du}{dx}, \text{where }u=1-8 x \text{and }\frac{d\text{}}{du}(\csc (u))=-\cot (u) \csc (u): \\
  \text{= }-3 e^{3/2+3 x} \csc (1-8 x)-\fbox{$-\cot (1-8 x) \csc (1-8 x) \left(\frac{d}{dx}(1-8 x)\right)$} e^{3/2+3 x} \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Simplify }\text{the }\text{expression}: \\
  \text{= }-3 e^{3/2+3 x} \csc (1-8 x)+e^{3/2+3 x} \cot (1-8 x) \csc (1-8 x) \left(\frac{d}{dx}(1-8 x)\right) \\
\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{= }-3 e^{3/2+3 x} \csc (1-8 x)+\fbox{$\frac{d}{dx}(1)-8 \left(\frac{d}{dx}(x)\right)$} e^{3/2+3 x} \cot (1-8 x) \csc (1-8 x) \\
\end{array}
 \\
 
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 \text{The }\text{derivative }\text{of }1 \text{is }\text{zero}: \\
  \text{= }-3 e^{3/2+3 x} \csc (1-8 x)+e^{3/2+3 x} \cot (1-8 x) \csc (1-8 x) \left(-8 \left(\frac{d}{dx}(x)\right)+\fbox{$0$}\right) \\
\end{array}
 \\
 
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 \text{Simplify }\text{the }\text{expression}: \\
  \text{= }-3 e^{3/2+3 x} \csc (1-8 x)-8 e^{3/2+3 x} \cot (1-8 x) \csc (1-8 x) \left(\frac{d}{dx}(x)\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{The }\text{derivative }\text{of }x \text{is }1: \\
  \text{= }-3 e^{3/2+3 x} \csc (1-8 x)-\fbox{$1$} 8 e^{3/2+3 x} \cot (1-8 x) \csc (1-8 x) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Simplify }\text{the }\text{expression}: \\
  \text{= }-3 e^{3/2+3 x} \csc (1-8 x)-8 e^{3/2+3 x} \cot (1-8 x) \csc (1-8 x) \\
\end{array}
 \\
 
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 \text{Simplify }\text{the }\text{expression}: \\
 \fbox{$
\begin{array}{ll}
 \text{Answer:} &  \\
 \text{} &  \text{= }-e^{3/2+3 x} (3+8 \cot (1-8 x)) \csc (1-8 x) \\
\end{array}
$} \\
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 \\
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{'source': 'AMPS', 'file': 'amps/mathematica/calculus/derivatives_w_steps/1994.txt'}