a)
\(y' = (2x{e^x})' + 3(\sin 2x)' \)
\(= 2.{e^x} + 2x({e^x})'+ {\rm{ }}3.2cos2x\)
\(=2\left( {1 + x} \right){e^x} + 6cos2x\)
b)
\(\begin{array}{l}y' = 5.2x - \left( {\left( {{2^x}} \right)'.\cos x + {2^x}.\left( {\cos x} \right)'} \right)\\\,\,\,\,\,\, = 10x - \left( {{2^x}.\ln 2.\cos x - {2^x}.\sin x} \right)\\\,\,\,\,\,\, = 10x - {2^x}\left( {\ln 2\cos x - \sin x} \right)\end{array}\)
c)
\(\begin{array}{l}y' = \frac{{\left( {x + 1} \right)'{{.3}^x} - \left( {x + 1} \right).\left( {{3^x}} \right)'}}{{{{\left( {{3^x}} \right)}^2}}}\\\,\,\,\,\, = \frac{{{3^x} - \left( {x + 1} \right){{.3}^x}\ln 3}}{{{{\left( {{3^x}} \right)}^2}}}\\\,\,\,\, = \frac{{{3^x}\left( {1 - \left( {x + 1} \right)\ln 3} \right)}}{{{{\left( {{3^x}} \right)}^2}}}\\\,\,\,\, = \frac{{1 - \left( {x + 1} \right)\ln 3}}{{{3^x}}}\end{array}\)
