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

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

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(\partial)/(\partial x)(xyln(x^2+y^2))

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

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

\frac{d}{d x} (ln (cosh (3 x)))

x \to \frac{π}{4} lim (\frac{sin ( 2 x ) - 1}{4 x - π})

\frac{d}{d x} (cos (2 x + 3 y))

y^{^{''}} - 16 y^{^{'}} + 63 y = 240 e^{x}

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

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