What is the derivative of (x^2-4/(6x^3-2)) ?

The derivative of (x^2-4/(6x^3-2)) is -(x(3x^3-36x+2))/(2(3x^3-1)^2)

What is the first derivative of (x^2-4/(6x^3-2)) ?

The first derivative of (x^2-4/(6x^3-2)) is -(x(3x^3-36x+2))/(2(3x^3-1)^2)

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\frac{d}{d x} (\frac{x ^{2} - 4}{6 x ^{3} - 2})

- \frac{x ( 3 x ^{3} - 36 x + 2 )}{2 ( 3 x ^{3} - 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{2} - 4}{6 x ^{3} - 2})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{2} - 4 ) ( 6 x ^{3} - 2 ) - \frac{d}{d x} ( 6 x ^{3} - 2 ) ( x ^{2} - 4 )}{( 6 x ^{3} - 2 ) ^{2}}

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

\frac{d}{d x} (6 x^{3} - 2) = 18 x^{2}

= \frac{2 x ( 6 x ^{3} - 2 ) - 18 x ^{2} ( x ^{2} - 4 )}{( 6 x ^{3} - 2 ) ^{2}}

Simplify

\frac{2 x ( 6 x ^{3} - 2 ) - 18 x ^{2} ( x ^{2} - 4 )}{( 6 x ^{3} - 2 ) ^{2}} : - \frac{x ( 3 x ^{3} - 36 x + 2 )}{2 ( 3 x ^{3} - 1 ) ^{2}}

= - \frac{x ( 3 x ^{3} - 36 x + 2 )}{2 ( 3 x ^{3} - 1 ) ^{2}}

\frac{1}{2 x} + \frac{1}{4 x ^{\frac{3}{4}}}

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

\frac{d}{d x} (arcsin (2 x - 1))

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 e^{2 t}, y (0) = 1, y^{^{'}} (0) = 0

\frac{d}{d x} (30 e^{x})

What is the derivative of (x^2-4/(6x^3-2)) ?
The derivative of (x^2-4/(6x^3-2)) is -(x(3x^3-36x+2))/(2(3x^3-1)^2)
What is the first derivative of (x^2-4/(6x^3-2)) ?
The first derivative of (x^2-4/(6x^3-2)) is -(x(3x^3-36x+2))/(2(3x^3-1)^2)