What is the derivative of (2x+3x^2/(x^2e^x)) ?

The derivative of (2x+3x^2/(x^2e^x)) is -(3x^2+2x+2)/(e^xx^2)

What is the first derivative of (2x+3x^2/(x^2e^x)) ?

The first derivative of (2x+3x^2/(x^2e^x)) is -(3x^2+2x+2)/(e^xx^2)

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\frac{d}{d x} (\frac{2 x + 3 x ^{2}}{x ^{2} e ^{x}})

- \frac{3 x ^{2} + 2 x + 2}{e ^{x} x ^{2}}

Solution steps

\frac{d}{d x} (\frac{2 x + 3 x ^{2}}{x ^{2} e ^{x}})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( 2 x + 3 x ^{2} ) x ^{2} e ^{x} - \frac{d}{d x} ( x ^{2} e ^{x} ) ( 2 x + 3 x ^{2} )}{( x ^{2} e ^{x} ) ^{2}}

\frac{d}{d x} (2 x + 3 x^{2}) = 2 + 6 x

\frac{d}{d x} (x^{2} e^{x}) = 2 x e^{x} + e^{x} x^{2}

= \frac{( 2 + 6 x ) x ^{2} e ^{x} - ( 2 x e ^{x} + e ^{x} x ^{2} ) ( 2 x + 3 x ^{2} )}{( x ^{2} e ^{x} ) ^{2}}

Simplify

\frac{( 2 + 6 x ) x ^{2} e ^{x} - ( 2 x e ^{x} + e ^{x} x ^{2} ) ( 2 x + 3 x ^{2} )}{( x ^{2} e ^{x} ) ^{2}} : - \frac{3 x ^{2} + 2 x + 2}{e ^{x} x ^{2}}

= - \frac{3 x ^{2} + 2 x + 2}{e ^{x} x ^{2}}

What is the derivative of (2x+3x^2/(x^2e^x)) ?
The derivative of (2x+3x^2/(x^2e^x)) is -(3x^2+2x+2)/(e^xx^2)
What is the first derivative of (2x+3x^2/(x^2e^x)) ?
The first derivative of (2x+3x^2/(x^2e^x)) is -(3x^2+2x+2)/(e^xx^2)