What is the derivative of bxe^{2xy} ?

The derivative of bxe^{2xy} is b(e^{2xy}+2e^{2xy}x(y+x(dy)/(dx)))

What is the first derivative of bxe^{2xy} ?

The first derivative of bxe^{2xy} is b(e^{2xy}+2e^{2xy}x(y+x(dy)/(dx)))

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^{2 x y})

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

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

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

= b (1 \cdot e^{2 x y} + e^{2 x y} \cdot 2 (y + x \frac{d y}{d x}) x)

Simplify

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

\int 1 + \frac{9 x}{4} d x

\frac{\partial}{\partial y} (ln (5 x + 4 y))

\frac{\partial}{\partial x} (2 x y - y e^{\frac{y}{z}})

4 e^{x} + x

\int_{0}^{9} \frac{1}{81 + t ^{2}} d t

What is the derivative of bxe^{2xy} ?
The derivative of bxe^{2xy} is b(e^{2xy}+2e^{2xy}x(y+x(dy)/(dx)))
What is the first derivative of bxe^{2xy} ?
The first derivative of bxe^{2xy} is b(e^{2xy}+2e^{2xy}x(y+x(dy)/(dx)))