What is (\partial)/(\partial x)(x^2y^3(6-x-y)) ?

The answer to (\partial)/(\partial x)(x^2y^3(6-x-y)) is y^3(-3x^2-2xy+12x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial x)(x^2y^3(6-x-y))

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

y^{3} (- 3 x^{2} - 2 x y + 12 x)

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

f = x^{2}, g = 6 - x - y

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

\frac{\partial}{\partial x} (x^{2}) = 2 x

\frac{\partial}{\partial x} (6 - x - y) = - 1

= y^{3} (2 x (6 - x - y) + (- 1) x^{2})

Simplify

y^{3} (2 x (6 - x - y) + (- 1) x^{2}) : y^{3} (- 3 x^{2} - 2 x y + 12 x)

= y^{3} (- 3 x^{2} - 2 x y + 12 x)

\frac{d}{d x} (x^{15})

y^{^{'}} = sin (x) y + e^{x}

\frac{\partial}{\partial θ} (\frac{cos ( θ )}{r ( θ )})

x \to 2 lim (\frac{0}{( x - 2 ) ^{2}})

\int 4 e^{6} d x

What is (\partial)/(\partial x)(x^2y^3(6-x-y)) ?
The answer to (\partial)/(\partial x)(x^2y^3(6-x-y)) is y^3(-3x^2-2xy+12x)