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

The derivative of (4-3x-x^2/(x^2-6)) is (3x^2+4x+18)/((x^2-6)^2)

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

The first derivative of (4-3x-x^2/(x^2-6)) is (3x^2+4x+18)/((x^2-6)^2)

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Expand

(- 2 x - 3) (x^{2} - 6) - 2 x (4 - 3 x - x^{2}) : 3 x^{2} + 4 x + 18

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

n = 0 \sum \infty \frac{1}{( n + 1 ) !}

f (x) = 2 x cos (x)

f (x) = 5 x^{2} - 6 x + 1, at x = 2

(sin (θ)) \frac{d y}{d θ} = cos (θ)

\int (x^{3} - 1) e^{- \frac{x}{3}} d x

What is the derivative of (4-3x-x^2/(x^2-6)) ?
The derivative of (4-3x-x^2/(x^2-6)) is (3x^2+4x+18)/((x^2-6)^2)
What is the first derivative of (4-3x-x^2/(x^2-6)) ?
The first derivative of (4-3x-x^2/(x^2-6)) is (3x^2+4x+18)/((x^2-6)^2)