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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

(e^{- x^{2}})^{'} = - 2 e^{- x^{2}} 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}}

\frac{d ^{2} y}{d t ^{2}} + 2 \frac{d y}{d t} + 2 y = cos (2 t)

7 e

x \cdot \frac{d y}{d x} = 7 x e^{x} - y + 9 x^{2}

\int e^{3 θ} sin (2 θ) d θ

\int 1 + 4 θ^{2} d θ

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