What is (\partial)/(\partial x)(ln(x^2-4xy+y^2)) ?

The answer to (\partial)/(\partial x)(ln(x^2-4xy+y^2)) is (2x-4y)/(x^2-4xy+y^2)

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(\partial)/(\partial x)(ln(x^2-4xy+y^2))

\frac{\partial}{\partial x} (ln (x^{2} - 4 x y + y^{2}))

\frac{2 x - 4 y}{x ^{2} - 4 x y + y ^{2}}

Solution steps

\frac{\partial}{\partial x} (ln (x^{2} - 4 x y + y^{2}))

Treat

y

as a constant

Apply the chain rule:

\frac{1}{x ^{2} - 4 x y + y ^{2}} \frac{\partial}{\partial x} (x^{2} - 4 x y + y^{2})

= \frac{1}{x ^{2} - 4 x y + y ^{2}} \frac{\partial}{\partial x} (x^{2} - 4 x y + y^{2})

\frac{\partial}{\partial x} (x^{2} - 4 x y + y^{2}) = 2 x - 4 y

= \frac{1}{x ^{2} - 4 x y + y ^{2}} (2 x - 4 y)

Simplify

\frac{1}{x ^{2} - 4 x y + y ^{2}} (2 x - 4 y) : \frac{2 x - 4 y}{x ^{2} - 4 x y + y ^{2}}

= \frac{2 x - 4 y}{x ^{2} - 4 x y + y ^{2}}

f (x) = (ln (5))^{t}

x \to 0 lim (\frac{4}{17 + x})

\int \frac{1}{( x ^{2} + 9 )} d x

\frac{d}{d x} (x^{2} + x)

\frac{d}{d x} ((\frac{1}{x ^{2}} - \frac{2}{x ^{4}}) (x + 8 x^{3}))

What is (\partial)/(\partial x)(ln(x^2-4xy+y^2)) ?
The answer to (\partial)/(\partial x)(ln(x^2-4xy+y^2)) is (2x-4y)/(x^2-4xy+y^2)