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

The derivative of (4x^4-2x/(4x^4+2x)) is (12x^2)/((2x^3+1)^2)

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

The first derivative of (4x^4-2x/(4x^4+2x)) is (12x^2)/((2x^3+1)^2)

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

\frac{12 x ^{2}}{( 2 x ^{3} + 1 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (4 x^{4} - 2 x) = 16 x^{3} - 2

\frac{d}{d x} (4 x^{4} + 2 x) = 16 x^{3} + 2

= \frac{( 16 x ^{3} - 2 ) ( 4 x ^{4} + 2 x ) - ( 16 x ^{3} + 2 ) ( 4 x ^{4} - 2 x )}{( 4 x ^{4} + 2 x ) ^{2}}

Simplify

\frac{( 16 x ^{3} - 2 ) ( 4 x ^{4} + 2 x ) - ( 16 x ^{3} + 2 ) ( 4 x ^{4} - 2 x )}{( 4 x ^{4} + 2 x ) ^{2}} : \frac{12 x ^{2}}{( 2 x ^{3} + 1 ) ^{2}}

= \frac{12 x ^{2}}{( 2 x ^{3} + 1 ) ^{2}}

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

4 x (x^{2} - 4)

\frac{\partial}{\partial x} (7 y - y^{3} - 4 x^{2} y)

y^{5} y^{^{'}} = - 2 cos (π x)

\int ln (3 x) d x

What is the derivative of (4x^4-2x/(4x^4+2x)) ?
The derivative of (4x^4-2x/(4x^4+2x)) is (12x^2)/((2x^3+1)^2)
What is the first derivative of (4x^4-2x/(4x^4+2x)) ?
The first derivative of (4x^4-2x/(4x^4+2x)) is (12x^2)/((2x^3+1)^2)