Problem:
Differentiate the following function:
$-\cos ^{-1}(8-5 x)-\tan (8 x+6)$
Answer:
\begin{array}{l}
 
\begin{array}{l}
 \text{Possible derivation:} \\
 \frac{d}{dx}\left(-\cos ^{-1}(8-5 x)-\tan (6+8 x)\right) \\
\end{array}
 \\
\hline
 
\begin{array}{l}
 \text{Differentiate }\text{the }\text{sum }\text{term }\text{by }\text{term }\text{and }\text{factor }\text{out }\text{constants:} \\
  \text{= }-\left(\frac{d}{dx}\left(\cos ^{-1}(8-5 x)\right)\right)-\frac{d}{dx}(\tan (6+8 x)) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Using }\text{the }\text{chain }\text{rule, }\frac{d}{dx}\left(\cos ^{-1}(8-5 x)\right)=\frac{d\cos ^{-1}(u)}{du} \frac{du}{dx}, \text{where }u=8-5 x \text{and }\frac{d\text{}}{du}\left(\cos ^{-1}(u)\right)=-\frac{1}{\sqrt{1-u^2}}: \\
  \text{= }-\left(\frac{d}{dx}(\tan (6+8 x))\right)-\fbox{$-\frac{\frac{d}{dx}(8-5 x)}{\sqrt{1-(8-5 x)^2}}$} \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Simplify }\text{the }\text{expression}: \\
  \text{= }\frac{\frac{d}{dx}(8-5 x)}{\sqrt{1-(8-5 x)^2}}-\frac{d}{dx}(\tan (6+8 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{= }-\left(\frac{d}{dx}(\tan (6+8 x))\right)+\fbox{$\frac{d}{dx}(8)-5 \left(\frac{d}{dx}(x)\right)$} \frac{1}{\sqrt{1-(8-5 x)^2}} \\
\end{array}
 \\
 
\begin{array}{l}
 \text{The }\text{derivative }\text{of }8 \text{is }\text{zero}: \\
  \text{= }-\left(\frac{d}{dx}(\tan (6+8 x))\right)+\frac{-5 \left(\frac{d}{dx}(x)\right)+\fbox{$0$}}{\sqrt{1-(8-5 x)^2}} \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Simplify }\text{the }\text{expression}: \\
  \text{= }-\frac{5 \left(\frac{d}{dx}(x)\right)}{\sqrt{1-(8-5 x)^2}}-\frac{d}{dx}(\tan (6+8 x)) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{The }\text{derivative }\text{of }x \text{is }1: \\
  \text{= }-\left(\frac{d}{dx}(\tan (6+8 x))\right)-\fbox{$1$} \frac{5}{\sqrt{1-(8-5 x)^2}} \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Using }\text{the }\text{chain }\text{rule, }\frac{d}{dx}(\tan (8 x+6))=\frac{d\tan (u)}{du} \frac{du}{dx}, \text{where }u=8 x+6 \text{and }\frac{d\text{}}{du}(\tan (u))=\sec ^2(u): \\
  \text{= }-\frac{5}{\sqrt{1-(8-5 x)^2}}-\fbox{$\left(\frac{d}{dx}(6+8 x)\right) \sec ^2(6+8 x)$} \\
\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{5}{\sqrt{1-(8-5 x)^2}}-\fbox{$\frac{d}{dx}(6)+8 \left(\frac{d}{dx}(x)\right)$} \sec ^2(6+8 x) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{The }\text{derivative }\text{of }6 \text{is }\text{zero}: \\
  \text{= }-\frac{5}{\sqrt{1-(8-5 x)^2}}-\sec ^2(6+8 x) \left(8 \left(\frac{d}{dx}(x)\right)+\fbox{$0$}\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Simplify }\text{the }\text{expression}: \\
  \text{= }-\frac{5}{\sqrt{1-(8-5 x)^2}}-8 \left(\frac{d}{dx}(x)\right) \sec ^2(6+8 x) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{The }\text{derivative }\text{of }x \text{is }1: \\
  \text{= }-\frac{5}{\sqrt{1-(8-5 x)^2}}-\fbox{$1$} 8 \sec ^2(6+8 x) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Simplify }\text{the }\text{expression}: \\
 \fbox{$
\begin{array}{ll}
 \text{Answer:} &  \\
 \text{} &  \text{= }-\frac{5}{\sqrt{1-(8-5 x)^2}}-8 \sec ^2(6+8 x) \\
\end{array}
$} \\
\end{array}
 \\
\end{array}

{'source': 'AMPS', 'file': 'amps/mathematica/calculus/derivatives_w_steps/7114.txt'}