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

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

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^2+xy))

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{2000}{( 2000 - P ) P} P^{^{'}} = 0.05

30

y^{^{''}} + y^{^{'}} - 5 x = 0

\int_{- \infty}^{x} \frac{1}{x ^{2}} d x

\frac{10}{s ^{2} + 8 s + 25}

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