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

The derivative of y=((x^2+1))/((x^3-1)) is (-x^4-3x^2-2x)/((x^3-1)^2)

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

The first derivative of y=((x^2+1))/((x^3-1)) is (-x^4-3x^2-2x)/((x^3-1)^2)

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derivative of

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

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

Solution steps

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

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

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

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

Expand

2 x (x^{3} - 1) - 3 x^{2} (x^{2} + 1) : - x^{4} - 3 x^{2} - 2 x

= \frac{- x ^{4} - 3 x ^{2} - 2 x}{( x ^{3} - 1 ) ^{2}}

\int sec (t) (8 sec (t) + 5 tan (t)) d t

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

x \to 0 lim ((x)^{3})

\frac{\partial}{\partial x} (3 sin (2 x) cos (6 y))

\frac{\partial}{\partial x} (ln (1 - x))

What is the derivative of y=((x^2+1))/((x^3-1)) ?
The derivative of y=((x^2+1))/((x^3-1)) is (-x^4-3x^2-2x)/((x^3-1)^2)
What is the first derivative of y=((x^2+1))/((x^3-1)) ?
The first derivative of y=((x^2+1))/((x^3-1)) is (-x^4-3x^2-2x)/((x^3-1)^2)