What is the derivative of f(x)=(2x^2+2x-4)/(x^2+x-6) ?

The derivative of f(x)=(2x^2+2x-4)/(x^2+x-6) is (-16x-8)/((x^2+x-6)^2)

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

The first derivative of f(x)=(2x^2+2x-4)/(x^2+x-6) is (-16x-8)/((x^2+x-6)^2)

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derivative of

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

\frac{- 16 x - 8}{( x ^{2} + x - 6 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Expand

(4 x + 2) (x^{2} + x - 6) - (2 x + 1) (2 x^{2} + 2 x - 4) : - 16 x - 8

= \frac{- 16 x - 8}{( x ^{2} + x - 6 ) ^{2}}

\frac{d}{d x} (2 (e^{- x} - e^{- x} x))

x \to \infty lim (\frac{x}{28})

\frac{\partial}{\partial x} (5 x^{3} y^{6})

\frac{\partial}{\partial x} (2 y - x)

6 - t

What is the derivative of f(x)=(2x^2+2x-4)/(x^2+x-6) ?
The derivative of f(x)=(2x^2+2x-4)/(x^2+x-6) is (-16x-8)/((x^2+x-6)^2)
What is the first derivative of f(x)=(2x^2+2x-4)/(x^2+x-6) ?
The first derivative of f(x)=(2x^2+2x-4)/(x^2+x-6) is (-16x-8)/((x^2+x-6)^2)