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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

x \to 0 + lim (x ln (x))

\frac{d}{d x} (sin (6 x) cos (4 x))

x^{4} + y^{3} x^{5} + 5 y^{^{'}} = 0

3 x^{2} y^{2} d x + 2 x^{3} y d y = 0

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