What is (\partial)/(\partial y)(x^2+y^2-1) ?

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

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^2-1)

\frac{\partial}{\partial y} (x^{2} + y^{2} - 1)

2 y

Solution steps

\frac{\partial}{\partial y} (x^{2} + y^{2} - 1)

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} (1)

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

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

\frac{\partial}{\partial y} (1) = 0

= 0 + 2 y - 0

Simplify

= 2 y

\frac{d}{d x} (- \frac{14}{x ^{6}})

4 (2 t - 5)^{2}

\int x^{2} \cdot x^{2} - 9 d x

\frac{d P}{d t} = 0.5 (P) (1 - \frac{P}{1000})

\frac{d y}{d x} = \frac{3 sec ( y )}{( x + 1 ) ^{2}}

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