What is the derivative of (3x/(2x^2+1)) ?

The derivative of (3x/(2x^2+1)) is (3(-2x^2+1))/((2x^2+1)^2)

What is the first derivative of (3x/(2x^2+1)) ?

The first derivative of (3x/(2x^2+1)) is (3(-2x^2+1))/((2x^2+1)^2)

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (2 x^{2} + 1) = 4 x

= 3 \cdot \frac{1 \cdot ( 2 x ^{2} + 1 ) - 4 xx}{( 2 x ^{2} + 1 ) ^{2}}

Simplify

3 \cdot \frac{1 \cdot ( 2 x ^{2} + 1 ) - 4 xx}{( 2 x ^{2} + 1 ) ^{2}} : \frac{3 ( - 2 x ^{2} + 1 )}{( 2 x ^{2} + 1 ) ^{2}}

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

\frac{\partial}{\partial x} (\frac{1}{2} (x^{2} + 6 y^{2}))

f (x) = x (x^{2} - 4 x + 5)^{4}, at x = 2

\frac{1}{s ^{2} + 2 s - 3}

\int \frac{e ^{- s t}}{s} d t

\int ln (6 x + 1) d x

What is the derivative of (3x/(2x^2+1)) ?
The derivative of (3x/(2x^2+1)) is (3(-2x^2+1))/((2x^2+1)^2)
What is the first derivative of (3x/(2x^2+1)) ?
The first derivative of (3x/(2x^2+1)) is (3(-2x^2+1))/((2x^2+1)^2)