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

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

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

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

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derivative of

f (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}}

\int 10 arcsin (x) d x

x \to - 2 lim ((x + 1) (x + 3))

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

y = 3 x^{2} + 12 x + 6

y^{^{''}} + 4 y^{^{'}} + 4 y = 8 x^{2}

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