What is the derivative of (e^{-x^2}/(x^3)) ?

The derivative of (e^{-x^2}/(x^3)) is -(e^{-x^2}(2x^2+3))/(x^4)

What is the first derivative of (e^{-x^2}/(x^3)) ?

The first derivative of (e^{-x^2}/(x^3)) is -(e^{-x^2}(2x^2+3))/(x^4)

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

- \frac{e ^{- x^{2}} ( 2 x ^{2} + 3 )}{x ^{4}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{( - 2 e ^{- x^{2}} x ) x ^{3} - 3 x ^{2} e ^{- x^{2}}}{( x ^{3} ) ^{2}} : - \frac{e ^{- x^{2}} ( 2 x ^{2} + 3 )}{x ^{4}}

= - \frac{e ^{- x^{2}} ( 2 x ^{2} + 3 )}{x ^{4}}

ln (6 - x^{2} + 2 x^{4})

\int_{1}^{\infty} \frac{7}{x ^{2} + 1} d x

y = ((\frac{x ^{2} - 2}{2 x ^{2} + 1})^{2})

\int 6 d x

\frac{d}{d x} (∣ 2 x - 1 ∣)

What is the derivative of (e^{-x^2}/(x^3)) ?
The derivative of (e^{-x^2}/(x^3)) is -(e^{-x^2}(2x^2+3))/(x^4)
What is the first derivative of (e^{-x^2}/(x^3)) ?
The first derivative of (e^{-x^2}/(x^3)) is -(e^{-x^2}(2x^2+3))/(x^4)