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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

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