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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

\frac{d x}{d x} = 1

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

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

Simplify

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

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

\frac{d}{d x} (\frac{5 x - 4}{3 x ^{2} + 1})

y^{^{''}} + 6 y^{^{'}} + 34 y = 0, y (0) = - 1, y^{^{'}} (0) = 13

f (x) = \frac{28}{x}

x \to 1 lim (2 x + 1)

x \to 0 - lim (\frac{e ^{\frac{1}{x}}}{x})

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