What is (\partial)/(\partial y)(x^2+y^2-5) ?

The answer to (\partial)/(\partial y)(x^2+y^2-5) 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^2-5)

\frac{\partial}{\partial y} (x^{2} + y^{2} - 5)

2 y

Solution steps

\frac{\partial}{\partial y} (x^{2} + y^{2} - 5)

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} (5)

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

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

\frac{\partial}{\partial y} (5) = 0

= 0 + 2 y - 0

Simplify

= 2 y

\int \frac{6}{5 ( 4 - 7 x ) ^{3}} d x

\int \frac{1}{1 + t ^{4}} d t

\int \frac{6 e ^{x}}{1 - e ^{2 x}} d x

\int 8 sec^{6} (t) d t

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

What is (\partial)/(\partial y)(x^2+y^2-5) ?
The answer to (\partial)/(\partial y)(x^2+y^2-5) is 2y