What is (\partial)/(\partial x)(x^2+y^2-4y) ?

The answer to (\partial)/(\partial x)(x^2+y^2-4y) is 2x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial x)(x^2+y^2-4y)

\frac{\partial}{\partial x} (x^{2} + y^{2} - 4 y)

2 x

Solution steps

\frac{\partial}{\partial x} (x^{2} + y^{2} - 4 y)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

\frac{\partial}{\partial x} (4 y) = 0

= 2 x + 0 - 0

Simplify

= 2 x

x \to 2 lim (5^{2} - 4 x + 1)

n = 0 \sum \infty 2 \cdot (\frac{4}{5})^{n}

y = 9 - x^{2}, y = - 8 x

\frac{2 x}{x ^{2} - 4}

What is (\partial)/(\partial x)(x^2+y^2-4y) ?
The answer to (\partial)/(\partial x)(x^2+y^2-4y) is 2x