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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

y^{^{''}} + 5 y^{^{'}} + 6 y = 0, y (0) = 2, y^{^{'}} (0) = 0

\frac{\frac{1}{s + 1} - 1}{s ^{2} - 1}

\int z^{2} (\frac{1}{2 π} e^{- \frac{1}{2} z^{2}}) d z

\int 3 sin^{2} (x) d x

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