What is the derivative of (x-2/(9x^2e^x)) ?

The derivative of (x-2/(9x^2e^x)) is (e^xx^2-(2e^xx+e^xx^2)(x-2))/(9e^{2x)x^4}

What is the first derivative of (x-2/(9x^2e^x)) ?

The first derivative of (x-2/(9x^2e^x)) is (e^xx^2-(2e^xx+e^xx^2)(x-2))/(9e^{2x)x^4}

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\frac{d}{d x} (\frac{x - 2}{9 x ^{2} e ^{x}})

\frac{e ^{x} x ^{2} - ( 2 e ^{x} x + e ^{x} x ^{2} ) ( x - 2 )}{9 e ^{2 x} x ^{4}}

Solution steps

\frac{d}{d x} (\frac{x - 2}{9 x ^{2} e ^{x}})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{9} \frac{d}{d x} (\frac{x - 2}{x ^{2} e ^{x}})

Apply the Quotient Rule:

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

= \frac{1}{9} \cdot \frac{\frac{d}{d x} ( x - 2 ) x ^{2} e ^{x} - \frac{d}{d x} ( x ^{2} e ^{x} ) ( x - 2 )}{( x ^{2} e ^{x} ) ^{2}}

\frac{d}{d x} (x - 2) = 1

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

= \frac{1}{9} \cdot \frac{1 \cdot x ^{2} e ^{x} - ( 2 x e ^{x} + e ^{x} x ^{2} ) ( x - 2 )}{( x ^{2} e ^{x} ) ^{2}}

Simplify

\frac{1}{9} \cdot \frac{1 \cdot x ^{2} e ^{x} - ( 2 x e ^{x} + e ^{x} x ^{2} ) ( x - 2 )}{( x ^{2} e ^{x} ) ^{2}} : \frac{e ^{x} x ^{2} - ( 2 e ^{x} x + e ^{x} x ^{2} ) ( x - 2 )}{9 e ^{2 x} x ^{4}}

= \frac{e ^{x} x ^{2} - ( 2 e ^{x} x + e ^{x} x ^{2} ) ( x - 2 )}{9 e ^{2 x} x ^{4}}

What is the derivative of (x-2/(9x^2e^x)) ?
The derivative of (x-2/(9x^2e^x)) is (e^xx^2-(2e^xx+e^xx^2)(x-2))/(9e^{2x)x^4}
What is the first derivative of (x-2/(9x^2e^x)) ?
The first derivative of (x-2/(9x^2e^x)) is (e^xx^2-(2e^xx+e^xx^2)(x-2))/(9e^{2x)x^4}