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

The derivative of (2x^3+x^2-2x/(x^4-2x^2-4)) is (-2x^6-2x^5+2x^4-28x^2-8x+8)/((x^4-2x^2-4)^2)

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

The first derivative of (2x^3+x^2-2x/(x^4-2x^2-4)) is (-2x^6-2x^5+2x^4-28x^2-8x+8)/((x^4-2x^2-4)^2)

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

\frac{- 2 x ^{6} - 2 x ^{5} + 2 x ^{4} - 28 x ^{2} - 8 x + 8}{( x ^{4} - 2 x ^{2} - 4 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Expand

(6 x^{2} + 2 x - 2) (x^{4} - 2 x^{2} - 4) - (4 x^{3} - 4 x) (2 x^{3} + x^{2} - 2 x) : - 2 x^{6} - 2 x^{5} + 2 x^{4} - 28 x^{2} - 8 x + 8

= \frac{- 2 x ^{6} - 2 x ^{5} + 2 x ^{4} - 28 x ^{2} - 8 x + 8}{( x ^{4} - 2 x ^{2} - 4 ) ^{2}}

y = 6 x - x^{2} (1.5)

\frac{d}{d x} (x^{2} lo g_{e} (2 x))

\frac{d y}{d x} + 4 tan (4 x) y = 3 cos (4 x)

(0, 1), (4, - 5)

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

What is the derivative of (2x^3+x^2-2x/(x^4-2x^2-4)) ?
The derivative of (2x^3+x^2-2x/(x^4-2x^2-4)) is (-2x^6-2x^5+2x^4-28x^2-8x+8)/((x^4-2x^2-4)^2)
What is the first derivative of (2x^3+x^2-2x/(x^4-2x^2-4)) ?
The first derivative of (2x^3+x^2-2x/(x^4-2x^2-4)) is (-2x^6-2x^5+2x^4-28x^2-8x+8)/((x^4-2x^2-4)^2)