What is the derivative of bxe^x ?

The derivative of bxe^x is b(e^x+e^xx)

What is the first derivative of bxe^x ?

The first derivative of bxe^x is b(e^x+e^xx)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

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

f = x, g = e^{x}

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

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

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

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

Simplify

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

\frac{\partial}{\partial x} (\frac{2 x ^{2} y}{5 + x ^{2} y ^{2}})

\frac{d}{d x} (((x^{2} + 3)^{5} + x)^{2})

ln (5 + x)

\frac{\partial}{\partial x} (8 y^{5} + 8 y^{2} x^{7})

\frac{d}{d x} (\frac{20}{x + 1})

What is the derivative of bxe^x ?
The derivative of bxe^x is b(e^x+e^xx)
What is the first derivative of bxe^x ?
The first derivative of bxe^x is b(e^x+e^xx)