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

The derivative of (-2x^2+x+4/(x^2+4x-1)) is (-9x^2-4x-17)/((x^2+4x-1)^2)

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

The first derivative of (-2x^2+x+4/(x^2+4x-1)) is (-9x^2-4x-17)/((x^2+4x-1)^2)

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

\frac{- 9 x ^{2} - 4 x - 17}{( x ^{2} + 4 x - 1 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Expand

(- 4 x + 1) (x^{2} + 4 x - 1) - (2 x + 4) (- 2 x^{2} + x + 4) : - 9 x^{2} - 4 x - 17

= \frac{- 9 x ^{2} - 4 x - 17}{( x ^{2} + 4 x - 1 ) ^{2}}

x \to 0 lim (\frac{- x ^{3}}{x ^{3}})

\int \frac{x}{9 - 2 x ^{4}} d x

\int \frac{2 x + 1}{x ^{3} - 3 x ^{2} + 3 x - 1} d x

\frac{d y}{d x} x + y = \frac{1}{y ^{2}}

\int \frac{x ^{3} - x + 3}{x ^{2} + x - 2} d x

What is the derivative of (-2x^2+x+4/(x^2+4x-1)) ?
The derivative of (-2x^2+x+4/(x^2+4x-1)) is (-9x^2-4x-17)/((x^2+4x-1)^2)
What is the first derivative of (-2x^2+x+4/(x^2+4x-1)) ?
The first derivative of (-2x^2+x+4/(x^2+4x-1)) is (-9x^2-4x-17)/((x^2+4x-1)^2)