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

The derivative of (2xy)/((x^2+y^2)) is (2y(-x^2+y^2))/((x^2+y^2)^2)

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

The first derivative of (2xy)/((x^2+y^2)) is (2y(-x^2+y^2))/((x^2+y^2)^2)

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derivative of

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

\frac{d x}{d x} = 1

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

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

Simplify

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

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

\frac{d}{d x} (- (x + p)^{2} + a)

\frac{1}{( 1 - t ^{2} )}

5 e^{3} x

arcsin (3^{\frac{1}{2}}) x

(x^{3} + 4 x)^{3}

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