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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial y} (2 k y - 5 x)

\int y^{- \frac{1}{3}} d y

\int \frac{5 x - 12}{x ^{3} - 6 x ^{2} + 8 x} d x

\int \frac{x - 3}{x ^{2} + 1} d x

\frac{7}{2 x + 1}, at a = 4

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