What is (\partial)/(\partial x)(ln(x^2+3xy+2y^2)) ?

The answer to (\partial)/(\partial x)(ln(x^2+3xy+2y^2)) is (2x+3y)/(x^2+3xy+2y^2)

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(\partial)/(\partial x)(ln(x^2+3xy+2y^2))

\frac{\partial}{\partial x} (ln (x^{2} + 3 x y + 2 y^{2}))

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

Solution steps

\frac{\partial}{\partial x} (ln (x^{2} + 3 x y + 2 y^{2}))

Treat

y

as a constant

Apply the chain rule:

\frac{1}{x ^{2} + 3 x y + 2 y ^{2}} \frac{\partial}{\partial x} (x^{2} + 3 x y + 2 y^{2})

= \frac{1}{x ^{2} + 3 x y + 2 y ^{2}} \frac{\partial}{\partial x} (x^{2} + 3 x y + 2 y^{2})

\frac{\partial}{\partial x} (x^{2} + 3 x y + 2 y^{2}) = 2 x + 3 y

= \frac{1}{x ^{2} + 3 x y + 2 y ^{2}} (2 x + 3 y)

Simplify

\frac{1}{x ^{2} + 3 x y + 2 y ^{2}} (2 x + 3 y) : \frac{2 x + 3 y}{x ^{2} + 3 x y + 2 y ^{2}}

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

7 x^{2} y + \frac{d y}{d x} = 4 x^{2}

\frac{d}{d x} (y^{2} - 2 x y + 6 x)

k = 0 \sum \infty (1)^{k}

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

f (x) = 3 - (- 27)

What is (\partial)/(\partial x)(ln(x^2+3xy+2y^2)) ?
The answer to (\partial)/(\partial x)(ln(x^2+3xy+2y^2)) is (2x+3y)/(x^2+3xy+2y^2)