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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

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

\frac{d}{d x} ((x + 8)^{\frac{2}{3}})

\int \frac{3 ( x + 1 ) + 4}{x ( 2 x + 7 ) + 3} d x

x \to 0 lim (\frac{( x ^{2} - 3 x )}{x})

y^{^{'}} - 5 y x^{4} = 0

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