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

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

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+y))

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

\frac{1}{x ^{2} + y}

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

= \frac{1}{x ^{2} + y} \cdot 1

Simplify

= \frac{1}{x ^{2} + y}

x^{4} - 2 x^{2}, 2 x^{2}, [0, 2]

f (x) = \frac{8 x}{x + 7}, (2, 1.7777700000000 \dots)

y = 5^{x - 3}, (3, 1)

f (x) = x^{3} - 2 x^{2} + x + 1

\int_{- 2}^{- 1} \frac{1}{( x - 1 ) ^{3}} d x

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