What is the derivative of (6x/(2+x^2)) ?

The derivative of (6x/(2+x^2)) is (6(2-x^2))/((2+x^2)^2)

What is the first derivative of (6x/(2+x^2)) ?

The first derivative of (6x/(2+x^2)) is (6(2-x^2))/((2+x^2)^2)

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\frac{d}{d x} (\frac{6 x}{2 + x ^{2}})

\frac{6 ( 2 - x ^{2} )}{( 2 + x ^{2} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{6 x}{2 + x ^{2}})

Take the constant out:

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

= 6 \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}}

= 6 \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

= 6 \cdot \frac{1 \cdot ( 2 + x ^{2} ) - 2 xx}{( 2 + x ^{2} ) ^{2}}

Simplify

6 \cdot \frac{1 \cdot ( 2 + x ^{2} ) - 2 xx}{( 2 + x ^{2} ) ^{2}} : \frac{6 ( 2 - x ^{2} )}{( 2 + x ^{2} ) ^{2}}

= \frac{6 ( 2 - x ^{2} )}{( 2 + x ^{2} ) ^{2}}

x y^{^{'}} = 2 y + x^{3}

\frac{\partial}{\partial x} (k)

\int \frac{1}{a x ^{2} + b x + c} d x

y^{^{'}} - (\frac{2 x}{1 + x ^{2}}) y = 0, y (0) = 4

d er i v a t i v e (t^{2} - t) (t^{2} + t)

What is the derivative of (6x/(2+x^2)) ?
The derivative of (6x/(2+x^2)) is (6(2-x^2))/((2+x^2)^2)
What is the first derivative of (6x/(2+x^2)) ?
The first derivative of (6x/(2+x^2)) is (6(2-x^2))/((2+x^2)^2)