Problem:
Differentiate the following function:
$e^{4 x^3+4} \sin \left(1-5 x^5\right)$
Answer:
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 \text{Possible derivation:} \\
 \frac{d}{dx}\left(e^{4+4 x^3} \sin \left(1-5 x^5\right)\right) \\
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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^{4 x^3+4} \text{and }v=\sin \left(1-5 x^5\right): \\
  \text{= }e^{4+4 x^3} \left(\frac{d}{dx}\left(\sin \left(1-5 x^5\right)\right)\right)+\left(\frac{d}{dx}\left(e^{4+4 x^3}\right)\right) \sin \left(1-5 x^5\right) \\
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 \\
 
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 \text{Using }\text{the }\text{chain }\text{rule, }\frac{d}{dx}\left(\sin \left(1-5 x^5\right)\right)=\frac{d\sin (u)}{du} \frac{du}{dx}, \text{where }u=1-5 x^5 \text{and }\frac{d\text{}}{du}(\sin (u))=\cos (u): \\
  \text{= }\left(\frac{d}{dx}\left(e^{4+4 x^3}\right)\right) \sin \left(1-5 x^5\right)+\fbox{$\cos \left(1-5 x^5\right) \left(\frac{d}{dx}\left(1-5 x^5\right)\right)$} e^{4+4 x^3} \\
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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{= }\left(\frac{d}{dx}\left(e^{4+4 x^3}\right)\right) \sin \left(1-5 x^5\right)+\fbox{$\frac{d}{dx}(1)-5 \left(\frac{d}{dx}\left(x^5\right)\right)$} e^{4+4 x^3} \cos \left(1-5 x^5\right) \\
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 \\
 
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 \text{The }\text{derivative }\text{of }1 \text{is }\text{zero}: \\
  \text{= }\left(\frac{d}{dx}\left(e^{4+4 x^3}\right)\right) \sin \left(1-5 x^5\right)+e^{4+4 x^3} \cos \left(1-5 x^5\right) \left(-5 \left(\frac{d}{dx}\left(x^5\right)\right)+\fbox{$0$}\right) \\
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 \\
 
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 \text{Simplify }\text{the }\text{expression}: \\
  \text{= }-5 e^{4+4 x^3} \cos \left(1-5 x^5\right) \left(\frac{d}{dx}\left(x^5\right)\right)+\left(\frac{d}{dx}\left(e^{4+4 x^3}\right)\right) \sin \left(1-5 x^5\right) \\
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 \\
 
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 \text{Use }\text{the }\text{power }\text{rule, }\frac{d}{dx}\left(x^n\right)=n x^{n-1}, \text{where }n=5.\text{$\backslash $n}\frac{d}{dx}\left(x^5\right)=5 x^4: \\
  \text{= }\left(\frac{d}{dx}\left(e^{4+4 x^3}\right)\right) \sin \left(1-5 x^5\right)-\fbox{$5 x^4$} 5 e^{4+4 x^3} \cos \left(1-5 x^5\right) \\
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 \\
 
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 \text{Simplify }\text{the }\text{expression}: \\
  \text{= }-25 e^{4+4 x^3} x^4 \cos \left(1-5 x^5\right)+\left(\frac{d}{dx}\left(e^{4+4 x^3}\right)\right) \sin \left(1-5 x^5\right) \\
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 \\
 
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 \text{Using }\text{the }\text{chain }\text{rule, }\frac{d}{dx}\left(e^{4 x^3+4}\right)=\frac{de^u}{du} \frac{du}{dx}, \text{where }u=4 x^3+4 \text{and }\frac{d\text{}}{du}\left(e^u\right)=e^u: \\
  \text{= }-25 e^{4+4 x^3} x^4 \cos \left(1-5 x^5\right)+\fbox{$e^{4+4 x^3} \left(\frac{d}{dx}\left(4+4 x^3\right)\right)$} \sin \left(1-5 x^5\right) \\
\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{= }-25 e^{4+4 x^3} x^4 \cos \left(1-5 x^5\right)+\fbox{$\frac{d}{dx}(4)+4 \left(\frac{d}{dx}\left(x^3\right)\right)$} e^{4+4 x^3} \sin \left(1-5 x^5\right) \\
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 \\
 
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 \text{The }\text{derivative }\text{of }4 \text{is }\text{zero}: \\
  \text{= }-25 e^{4+4 x^3} x^4 \cos \left(1-5 x^5\right)+e^{4+4 x^3} \sin \left(1-5 x^5\right) \left(4 \left(\frac{d}{dx}\left(x^3\right)\right)+\fbox{$0$}\right) \\
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 \\
 
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 \text{Simplify }\text{the }\text{expression}: \\
  \text{= }-25 e^{4+4 x^3} x^4 \cos \left(1-5 x^5\right)+4 e^{4+4 x^3} \left(\frac{d}{dx}\left(x^3\right)\right) \sin \left(1-5 x^5\right) \\
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 \\
 
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 \text{Use }\text{the }\text{power }\text{rule, }\frac{d}{dx}\left(x^n\right)=n x^{n-1}, \text{where }n=3.\text{$\backslash $n}\frac{d}{dx}\left(x^3\right)=3 x^2: \\
  \text{= }-25 e^{4+4 x^3} x^4 \cos \left(1-5 x^5\right)+\fbox{$3 x^2$} 4 e^{4+4 x^3} \sin \left(1-5 x^5\right) \\
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 \\
 
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 \text{Simplify }\text{the }\text{expression}: \\
  \text{= }-25 e^{4+4 x^3} x^4 \cos \left(1-5 x^5\right)+12 e^{4+4 x^3} x^2 \sin \left(1-5 x^5\right) \\
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 \\
 
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 \text{Simplify }\text{the }\text{expression}: \\
 \fbox{$
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 \text{Answer:} &  \\
 \text{} &  \text{= }-e^{4+4 x^3} x^2 \left(25 x^2 \cos \left(1-5 x^5\right)-12 \sin \left(1-5 x^5\right)\right) \\
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$} \\
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 \\
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{'source': 'AMPS', 'file': 'amps/mathematica/calculus/derivatives_w_steps/6818.txt'}