What is the derivative of (2xy/(1+x^2)) ?

The derivative of (2xy/(1+x^2)) is (2(x^3\frac{dy)/(dx)-x^2y+x(dy)/(dx)+y)}{(1+x^2)^2}

What is the first derivative of (2xy/(1+x^2)) ?

The first derivative of (2xy/(1+x^2)) is (2(x^3\frac{dy)/(dx)-x^2y+x(dy)/(dx)+y)}{(1+x^2)^2}

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

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

Solution steps

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

Take the constant out:

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

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

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

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

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

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

Simplify

2 \cdot \frac{( y + x \frac{d y}{d x} ) ( 1 + x ^{2} ) - 2 xx y}{( 1 + x ^{2} ) ^{2}} : \frac{2 ( x ^{3} \frac{d y}{d x} - x ^{2} y + x \frac{d y}{d x} + y )}{( 1 + x ^{2} ) ^{2}}

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

y^{^{'}} = 2 e^{2 x - y}, y (0) = 0

\frac{d}{d x} (x^{3} - x^{2} + 7 x - 1)

\int 2 \cdot cos (x) d x

h \to 0 lim (\frac{cos ( h ) - 1}{h})

x \to 1 lim (e^{- x})

What is the derivative of (2xy/(1+x^2)) ?
The derivative of (2xy/(1+x^2)) is (2(x^3\frac{dy)/(dx)-x^2y+x(dy)/(dx)+y)}{(1+x^2)^2}
What is the first derivative of (2xy/(1+x^2)) ?
The first derivative of (2xy/(1+x^2)) is (2(x^3\frac{dy)/(dx)-x^2y+x(dy)/(dx)+y)}{(1+x^2)^2}