Problem:
Differentiate the following function:
$e^{6-5 x} \cos (4 x+4)$
Answer:
\begin{array}{l}
 
\begin{array}{l}
 \text{Possible derivation:} \\
 \frac{d}{dx}\left(e^{6-5 x} \cos (4+4 x)\right) \\
\end{array}
 \\
\hline
 
\begin{array}{l}
 \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^{6-5 x} \text{and }v=\cos (4 x+4): \\
  \text{= }\cos (4+4 x) \left(\frac{d}{dx}\left(e^{6-5 x}\right)\right)+e^{6-5 x} \left(\frac{d}{dx}(\cos (4+4 x))\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Using }\text{the }\text{chain }\text{rule, }\frac{d}{dx}\left(e^{6-5 x}\right)=\frac{de^u}{du} \frac{du}{dx}, \text{where }u=6-5 x \text{and }\frac{d\text{}}{du}\left(e^u\right)=e^u: \\
  \text{= }e^{6-5 x} \left(\frac{d}{dx}(\cos (4+4 x))\right)+\fbox{$e^{6-5 x} \left(\frac{d}{dx}(6-5 x)\right)$} \cos (4+4 x) \\
\end{array}
 \\
 
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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^{6-5 x} \left(\frac{d}{dx}(\cos (4+4 x))\right)+\fbox{$\frac{d}{dx}(6)-5 \left(\frac{d}{dx}(x)\right)$} e^{6-5 x} \cos (4+4 x) \\
\end{array}
 \\
 
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 \text{The }\text{derivative }\text{of }6 \text{is }\text{zero}: \\
  \text{= }e^{6-5 x} \left(\frac{d}{dx}(\cos (4+4 x))\right)+e^{6-5 x} \cos (4+4 x) \left(-5 \left(\frac{d}{dx}(x)\right)+\fbox{$0$}\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Simplify }\text{the }\text{expression}: \\
  \text{= }-5 e^{6-5 x} \cos (4+4 x) \left(\frac{d}{dx}(x)\right)+e^{6-5 x} \left(\frac{d}{dx}(\cos (4+4 x))\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{The }\text{derivative }\text{of }x \text{is }1: \\
  \text{= }e^{6-5 x} \left(\frac{d}{dx}(\cos (4+4 x))\right)-\fbox{$1$} 5 e^{6-5 x} \cos (4+4 x) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Using }\text{the }\text{chain }\text{rule, }\frac{d}{dx}(\cos (4 x+4))=\frac{d\cos (u)}{du} \frac{du}{dx}, \text{where }u=4 x+4 \text{and }\frac{d\text{}}{du}(\cos (u))=-\sin (u): \\
  \text{= }-5 e^{6-5 x} \cos (4+4 x)+\fbox{$-\left(\frac{d}{dx}(4+4 x)\right) \sin (4+4 x)$} e^{6-5 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{= }-5 e^{6-5 x} \cos (4+4 x)-\fbox{$\frac{d}{dx}(4)+4 \left(\frac{d}{dx}(x)\right)$} e^{6-5 x} \sin (4+4 x) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{The }\text{derivative }\text{of }4 \text{is }\text{zero}: \\
  \text{= }-5 e^{6-5 x} \cos (4+4 x)-e^{6-5 x} \sin (4+4 x) \left(4 \left(\frac{d}{dx}(x)\right)+\fbox{$0$}\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Simplify }\text{the }\text{expression}: \\
  \text{= }-5 e^{6-5 x} \cos (4+4 x)-4 e^{6-5 x} \left(\frac{d}{dx}(x)\right) \sin (4+4 x) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{The }\text{derivative }\text{of }x \text{is }1: \\
  \text{= }-5 e^{6-5 x} \cos (4+4 x)-\fbox{$1$} 4 e^{6-5 x} \sin (4+4 x) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Simplify }\text{the }\text{expression}: \\
  \text{= }-5 e^{6-5 x} \cos (4+4 x)-4 e^{6-5 x} \sin (4+4 x) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Simplify }\text{the }\text{expression}: \\
 \fbox{$
\begin{array}{ll}
 \text{Answer:} &  \\
 \text{} &  \text{= }-e^{6-5 x} (5 \cos (4+4 x)+4 \sin (4+4 x)) \\
\end{array}
$} \\
\end{array}
 \\
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{'source': 'AMPS', 'file': 'amps/mathematica/calculus/derivatives_w_steps/5955.txt'}