What is the derivative of (2x/(6x^2+1)) ?

The derivative of (2x/(6x^2+1)) is (2(-6x^2+1))/((6x^2+1)^2)

What is the first derivative of (2x/(6x^2+1)) ?

The first derivative of (2x/(6x^2+1)) is (2(-6x^2+1))/((6x^2+1)^2)

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

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

Solution steps

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

Take the constant out:

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

= 2 \frac{d}{d x} (\frac{x}{6 x ^{2} + 1})

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (6 x^{2} + 1) = 12 x

= 2 \cdot \frac{1 \cdot ( 6 x ^{2} + 1 ) - 12 xx}{( 6 x ^{2} + 1 ) ^{2}}

Simplify

2 \cdot \frac{1 \cdot ( 6 x ^{2} + 1 ) - 12 xx}{( 6 x ^{2} + 1 ) ^{2}} : \frac{2 ( - 6 x ^{2} + 1 )}{( 6 x ^{2} + 1 ) ^{2}}

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

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

\frac{d}{d x} (x - 3 ln (x))

\frac{d}{d x} (e^{- x t})

x \to 3 + lim (\frac{[ x ]}{x})

y^{^{''}} - y = e^{x}

What is the derivative of (2x/(6x^2+1)) ?
The derivative of (2x/(6x^2+1)) is (2(-6x^2+1))/((6x^2+1)^2)
What is the first derivative of (2x/(6x^2+1)) ?
The first derivative of (2x/(6x^2+1)) is (2(-6x^2+1))/((6x^2+1)^2)