What is (\partial)/(\partial x)((xy)/(x+y+z)) ?

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

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(\partial)/(\partial x)((xy)/(x+y+z))

\frac{\partial}{\partial x} (\frac{x y}{x + y + z})

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

Solution steps

\frac{\partial}{\partial x} (\frac{x y}{x + y + z})

Treat

y, z

as constants

Take the constant out:

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

= y \frac{\partial}{\partial x} (\frac{x}{x + y + z})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Simplify

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

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

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

(y) \to (0) lim (3 x + 7 x y - 6 y)

x \to 1 lim (\frac{x}{∣ x ∣ - 1})

0.1 y^{^{'}} - 0.1 y = 0

f (x) = sin (x^{2})

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