What is the derivative of axe^{bx^2} ?

The derivative of axe^{bx^2} is a(e^{bx^2}+2be^{bx^2}x^2)

What is the first derivative of axe^{bx^2} ?

The first derivative of axe^{bx^2} is a(e^{bx^2}+2be^{bx^2}x^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

\frac{d}{d x} (a x e^{b x^{2}})

a (e^{b x^{2}} + 2 b e^{b x^{2}} x^{2})

Solution steps

\frac{d}{d x} (a x e^{b x^{2}})

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

\frac{d x}{d x} = 1

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

= a (1 \cdot e^{b x^{2}} + e^{b x^{2}} \cdot 2 b xx)

Simplify

a (1 \cdot e^{b x^{2}} + e^{b x^{2}} \cdot 2 b xx) : a (e^{b x^{2}} + 2 b e^{b x^{2}} x^{2})

= a (e^{b x^{2}} + 2 b e^{b x^{2}} x^{2})

\frac{\partial}{\partial x} (3 x^{2} + 6 x y)

\frac{d}{d x} (\frac{12 ( 3 x ^{3} - 16 )}{4 x - x ^{4}})

\int p (p + 7)^{8} d p

y^{^{''}} + 2 y^{^{'}} + y = 4 e^{- 1}

\frac{d}{d t} (tan (t + \frac{π}{4}))

What is the derivative of axe^{bx^2} ?
The derivative of axe^{bx^2} is a(e^{bx^2}+2be^{bx^2}x^2)
What is the first derivative of axe^{bx^2} ?
The first derivative of axe^{bx^2} is a(e^{bx^2}+2be^{bx^2}x^2)