What is (\partial)/(\partial x)(2x^2+y^2+2xy+yx+7y) ?

The answer to (\partial)/(\partial x)(2x^2+y^2+2xy+yx+7y) is 4x+3y

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial x)(2x^2+y^2+2xy+yx+7y)

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

4 x + 3 y

Solution steps

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

Simplify

2 x^{2} + y^{2} + 2 x y + y x + 7 y : 2 x^{2} + y^{2} + 3 x y + 7 y

= \frac{\partial}{\partial x} (2 x^{2} + y^{2} + 3 x y + 7 y)

Treat

y

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial x} (2 x^{2}) + \frac{\partial}{\partial x} (y^{2}) + \frac{\partial}{\partial x} (3 x y) + \frac{\partial}{\partial x} (7 y)

\frac{\partial}{\partial x} (2 x^{2}) = 4 x

\frac{\partial}{\partial x} (y^{2}) = 0

\frac{\partial}{\partial x} (3 x y) = 3 y

\frac{\partial}{\partial x} (7 y) = 0

= 4 x + 0 + 3 y + 0

Simplify

= 4 x + 3 y

\frac{\partial}{\partial v} (ln (u cos (v)))

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

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

x \to \infty lim (- 1)

x \to 4 lim (2 x)

What is (\partial)/(\partial x)(2x^2+y^2+2xy+yx+7y) ?
The answer to (\partial)/(\partial x)(2x^2+y^2+2xy+yx+7y) is 4x+3y