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

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

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

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

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derivative of

y = \frac{2 x}{x ^{2} - 1}

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

Solution steps

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

Take the constant out:

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

= 2 \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}}

= 2 \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

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

Simplify

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

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

y (θ) = e^{t a n (θ)}

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

x \to - 2 lim (\frac{- 5 x - 10}{x ^{3} + 2 x ^{2}})

\frac{6 x}{x ^{2} + 2}

\frac{d y}{d x} = \frac{1}{x 16 x ^{2} - 1}

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