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

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

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

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

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derivative of

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

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

Solution steps

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

Simplify

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

= \frac{d}{d x} (- \frac{x + 3}{x + 1})

Take the constant out:

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

= - \frac{d}{d x} (\frac{x + 3}{x + 1})

Apply the Quotient Rule:

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

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

\frac{d}{d x} (x + 3) = 1

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

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

Simplify

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

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

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

\frac{\partial}{\partial x} (2 x y + y^{2} + 1)

\int \frac{3}{x ^{2}} + 4 x d x

y^{2} d x - x y d y = x^{2} y d y

\frac{1}{2} x^{2}, x^{2} - 3 x

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