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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial x)(ln(x+y^2))

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int w w^{2} - ln (3) d w

x \to 0 lim (\frac{3 x}{x e ^{x}})

\int \frac{3 x ^{2} - 2 x + 5}{x ^{2}} d x

x^{3} \frac{d y}{d x} + y^{2} = x^{2} y

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