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Popular >

(xe^{-x^2})^'

Solution

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

Solution

e^{- x^{2}} - 2 e^{- x^{2}} x^{2}

Solution steps

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = e^{- x^{2}}

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

x^{'} = 1

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

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

Simplify

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

= e^{- x^{2}} - 2 e^{- x^{2}} x^{2}

Popular Examples

integral of ((x+3))/((x^2+6x)^{1/3)}

\int \frac{( x + 3 )}{( x ^{2} + 6 x ) ^{\frac{1}{3}}} d x

(dy)/(dx)=(3x^4+y^3)/(3xy^2)

\frac{d y}{d x} = \frac{3 x ^{4} + y ^{3}}{3 x y ^{2}}

(dy)/(dt)= 1/(y-2)

\frac{d y}{d t} = \frac{1}{y - 2}

integral from 0 to 1 of 4xye^{-(2x+y)}

\int_{0}^{1} 4 x y e^{- (2 x + y)} d y

sum from n=0 to infinity of (-5/6)^n

n = 0 \sum \infty (- \frac{5}{6})^{n}

Frequently Asked Questions (FAQ)

What is the (xe^{-x^2})^' ?
The (xe^{-x^2})^' is e^{-x^2}-2e^{-x^2}x^2
What is the first (xe^{-x^2})^' ?
The first (xe^{-x^2})^' is e^{-x^2}-2e^{-x^2}x^2