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

The answer to (\partial)/(\partial y)(x^2+y^2+x+y+xy) is x+2y+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

x + 2 y + 1

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

\frac{\partial}{\partial y} (y) = 1

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

= 0 + 2 y + 0 + 1 + x

Simplify

= x + 2 y + 1

f (θ) = 5 θ

f (x) = 7 x - 4 x^{2}, (- 1, - 11)

f (x) = \frac{5}{x - 7}, (8, 5)

y = x^{16}

What is (\partial)/(\partial y)(x^2+y^2+x+y+xy) ?
The answer to (\partial)/(\partial y)(x^2+y^2+x+y+xy) is x+2y+1