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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

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

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

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

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

Simplify

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

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

f (x) = ln (6 - x^{2})

\frac{\partial}{\partial x} (x cos (y) - y sin (x))

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

x \to 1 lim (\frac{( 5 - 5 x )}{1 - x})

x \frac{d y}{d x} - y = 0, y (a) = b

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