What is the derivative of (6x/(4x^2+1)) ?

The derivative of (6x/(4x^2+1)) is (6(-4x^2+1))/((4x^2+1)^2)

What is the first derivative of (6x/(4x^2+1)) ?

The first derivative of (6x/(4x^2+1)) is (6(-4x^2+1))/((4x^2+1)^2)

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\frac{d}{d x} (\frac{6 x}{4 x ^{2} + 1})

\frac{6 ( - 4 x ^{2} + 1 )}{( 4 x ^{2} + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{6 x}{4 x ^{2} + 1})

Take the constant out:

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

= 6 \frac{d}{d x} (\frac{x}{4 x ^{2} + 1})

Apply the Quotient Rule:

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

= 6 \cdot \frac{\frac{d x}{d x} ( 4 x ^{2} + 1 ) - \frac{d}{d x} ( 4 x ^{2} + 1 ) x}{( 4 x ^{2} + 1 ) ^{2}}

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

\frac{d}{d x} (4 x^{2} + 1) = 8 x

= 6 \cdot \frac{1 \cdot ( 4 x ^{2} + 1 ) - 8 xx}{( 4 x ^{2} + 1 ) ^{2}}

Simplify

6 \cdot \frac{1 \cdot ( 4 x ^{2} + 1 ) - 8 xx}{( 4 x ^{2} + 1 ) ^{2}} : \frac{6 ( - 4 x ^{2} + 1 )}{( 4 x ^{2} + 1 ) ^{2}}

= \frac{6 ( - 4 x ^{2} + 1 )}{( 4 x ^{2} + 1 ) ^{2}}

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

\int \frac{17}{( x - 1 ) ( x ^{2} + 16 )} d x

\frac{d}{d x} ((f (x))^{g (x)})

\frac{\partial}{\partial y} (cos (x^{2} y))

\int (sec (t) - 1)^{2} d t

What is the derivative of (6x/(4x^2+1)) ?
The derivative of (6x/(4x^2+1)) is (6(-4x^2+1))/((4x^2+1)^2)
What is the first derivative of (6x/(4x^2+1)) ?
The first derivative of (6x/(4x^2+1)) is (6(-4x^2+1))/((4x^2+1)^2)