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

The derivative of (4-2x-x^2/(x^2-6)) is (2x^2+4x+12)/((x^2-6)^2)

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

The first derivative of (4-2x-x^2/(x^2-6)) is (2x^2+4x+12)/((x^2-6)^2)

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

\frac{2 x ^{2} + 4 x + 12}{( x ^{2} - 6 ) ^{2}}

Solution steps

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

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

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

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

Expand

(- 2 x - 2) (x^{2} - 6) - 2 x (4 - 2 x - x^{2}) : 2 x^{2} + 4 x + 12

= \frac{2 x ^{2} + 4 x + 12}{( x ^{2} - 6 ) ^{2}}

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

x^{4} ln (x)

4 y^{^{''}} - 8 y^{^{'}} + 7 y = 0

y^{^{''}} + 64 y = sec^{2} (8 x)

\int cos (4 x + 1) d x

What is the derivative of (4-2x-x^2/(x^2-6)) ?
The derivative of (4-2x-x^2/(x^2-6)) is (2x^2+4x+12)/((x^2-6)^2)
What is the first derivative of (4-2x-x^2/(x^2-6)) ?
The first derivative of (4-2x-x^2/(x^2-6)) is (2x^2+4x+12)/((x^2-6)^2)