What is the derivative of (4-3x-x^2/(x^2-1)) ?

The derivative of (4-3x-x^2/(x^2-1)) is 3/((x+1)^2)

What is the first derivative of (4-3x-x^2/(x^2-1)) ?

The first derivative of (4-3x-x^2/(x^2-1)) is 3/((x+1)^2)

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\frac{d}{d x} (\frac{4 - 3 x - x ^{2}}{x ^{2} - 1})

\frac{3}{( x + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{4 - 3 x - x ^{2}}{x ^{2} - 1})

Simplify

\frac{4 - 3 x - x ^{2}}{x ^{2} - 1} : - \frac{x + 4}{x + 1}

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (x + 1) = 1

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

Simplify

- \frac{1 \cdot ( x + 1 ) - 1 \cdot ( x + 4 )}{( x + 1 ) ^{2}} : \frac{3}{( x + 1 ) ^{2}}

= \frac{3}{( x + 1 ) ^{2}}

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

\frac{d h}{d t} = 5 h

1 + x^{2} d y - 1 - y^{2} d x = 0

\int (13 x + 3) d x

What is the derivative of (4-3x-x^2/(x^2-1)) ?
The derivative of (4-3x-x^2/(x^2-1)) is 3/((x+1)^2)
What is the first derivative of (4-3x-x^2/(x^2-1)) ?
The first derivative of (4-3x-x^2/(x^2-1)) is 3/((x+1)^2)