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

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

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

2 y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial y} (ln (x^{2} + 1)) = 0

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

= 0 + 2 y

Simplify

= 2 y

arctan (0)

x \to 1 lim (e^{x^{3} - x})

\frac{d}{d x} (\frac{3 x - 2}{x ^{2}})

\int (x^{5} - 9 + \frac{6}{x ^{7}}) d x

\frac{d}{d x} (y + x) + x + y + 1 = 0

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