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Scrat [10]
3 years ago
6

Y= ln(x^3 e^2x) (dy)/(dx)​

Mathematics
2 answers:
Anna71 [15]3 years ago
8 0

Answer:

2 + 3/x.

Step-by-step explanation:

y = ln(x^3 . e^(2x)

dy/dx = 1 /(x^3 e^2x) * (2x^3e^(2x)  + 3x^2 e ^(2x))

= 2 + 3/x.

uysha [10]3 years ago
7 0

Answer:

y =  ln( {x}^{3} . {e}^{2x} )

divide by In :

\frac{y}{ ln}  =  \frac{ ln( {x}^{3}. {e}^{2x}  ) }{ ln }  \\  \\  {e}^{y}  =  {x}^{3} . {e}^{2x}

find dy/dx:

{d( {e}^{y}) } = ( {x}^{3} . {e}^{2x} )dx \\  {e}^{y} dy =  \{(3 {x}^{2} . {e}^{2x} ) + ( {x}^{3} .2 {e}^{2x} ) \}dx \\  \\ {e}^{y}   \frac{dy}{dx}  =  {x}^{3}  + 3 {x}^{2}  + 3 {e}^{2x}  \\  \\  \frac{dy}{dx}  =  \frac{ {x}^{3} + 3 {x}^{2} + 3 {e}^{2x}   }{ {e}^{y} }  \\  \\  { \boxed{ \boxed{\frac{dy}{dx}  =  \frac{ {x}^{3} + 3 {x}^{2}   + 3 {e}^{2x} }{ {e}^{ {x}^{3}. {e}^{2x}  } }}}}

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