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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{2} - y^{2}, g = ln (x y)

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

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

\frac{\partial}{\partial x} (ln (x y)) = \frac{1}{x}

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

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

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

\frac{d}{d x} (9 sin^{2} (x))

y = x^{2} - 14, y = 13 x

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

\frac{10}{( s ( s + 1 ) )}

sin^{2} (y) d x + x sin (2 y) d y = 0

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