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can you help me with this for my exercises: I need to to show that: g(x)= (e(-3x+2sinx)/6) ---------------- (3+2cosx) has derivative g'(x)= - (4(sin^2)x -12sin x +5) (e(-3x+2sinx)/6) ----------------------------------- 6(3+2cosx)^2
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quotient rule !!
\[ g(x)=\frac{e^{\frac {1}{6} (2 \sin (x)-3 x)}}{\cos (x)+3}\\ g'(x)=\frac{e^{\frac{1}{6} (2 \sin (x)-3 x)} \left(6 \sin (x)+2 \cos ^2(x)+3 \cos (x)-9\right)}{6 (\cos (x)+3)^2} \]
\[ g'(x)=\frac{e^{\frac{1}{6} (2 \sin (x)-3 x)} \left(-2 \sin ^2(x)+6 \sin (x)+3 \cos (x)-7\right)}{6 (\cos (x)+3)^2} \]
It seems to me that your g'(x) is not 100% right.
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