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

The derivative of f(x)=(x^3+1)/(x^3-1) is -(6x^2)/((x^3-1)^2)

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

The first derivative of f(x)=(x^3+1)/(x^3-1) is -(6x^2)/((x^3-1)^2)

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derivative of

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

- \frac{6 x ^{2}}{( x ^{3} - 1 ) ^{2}}

Solution steps

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

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

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

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

Simplify

\frac{3 x ^{2} ( x ^{3} - 1 ) - 3 x ^{2} ( x ^{3} + 1 )}{( x ^{3} - 1 ) ^{2}} : - \frac{6 x ^{2}}{( x ^{3} - 1 ) ^{2}}

= - \frac{6 x ^{2}}{( x ^{3} - 1 ) ^{2}}

x^{3} y^{^{'''}} + x y^{^{'}} - y = 0

\int e^{- 1}

f (x) = \frac{2 ln ( x )}{x}

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

\frac{d}{d x} (- 2 (e^{x} sin (x) + cos (x) e^{x}))

What is the derivative of f(x)=(x^3+1)/(x^3-1) ?
The derivative of f(x)=(x^3+1)/(x^3-1) is -(6x^2)/((x^3-1)^2)
What is the first derivative of f(x)=(x^3+1)/(x^3-1) ?
The first derivative of f(x)=(x^3+1)/(x^3-1) is -(6x^2)/((x^3-1)^2)