What is the derivative of (2xy/(x+y)) ?

The derivative of (2xy/(x+y)) is (2(x^2\frac{dy)/(dx)+y^2)}{(x+y)^2}

What is the first derivative of (2xy/(x+y)) ?

The first derivative of (2xy/(x+y)) is (2(x^2\frac{dy)/(dx)+y^2)}{(x+y)^2}

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{d}{d x} (154)

(\frac{4 x ^{2}}{4 - x})^{3}

\frac{d y}{d x} = \frac{x - y}{x + y}

(\frac{1}{y ^{2}} - \frac{3}{y ^{4}}) (y + 5 y^{3})

x \to \infty lim (y)

What is the derivative of (2xy/(x+y)) ?
The derivative of (2xy/(x+y)) is (2(x^2\frac{dy)/(dx)+y^2)}{(x+y)^2}
What is the first derivative of (2xy/(x+y)) ?
The first derivative of (2xy/(x+y)) is (2(x^2\frac{dy)/(dx)+y^2)}{(x+y)^2}