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

The derivative of (x^2+3/(x^2-1)) is -(8x)/((x^2-1)^2)

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

The first derivative of (x^2+3/(x^2-1)) is -(8x)/((x^2-1)^2)

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

- \frac{8 x}{( x ^{2} - 1 ) ^{2}}

Solution steps

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

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

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

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

Simplify

\frac{2 x ( x ^{2} - 1 ) - 2 x ( x ^{2} + 3 )}{( x ^{2} - 1 ) ^{2}} : - \frac{8 x}{( x ^{2} - 1 ) ^{2}}

= - \frac{8 x}{( x ^{2} - 1 ) ^{2}}

\frac{sin ( 4 t )}{cos ( 6 t )}

y = x^{3 x}

y^{^{''}} + y = sec (t)

n = 1 \sum \infty \frac{6 n}{9 - 2 n}

f (x) = (1 + x)^{- \frac{5}{3}}

What is the derivative of (x^2+3/(x^2-1)) ?
The derivative of (x^2+3/(x^2-1)) is -(8x)/((x^2-1)^2)
What is the first derivative of (x^2+3/(x^2-1)) ?
The first derivative of (x^2+3/(x^2-1)) is -(8x)/((x^2-1)^2)