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

The derivative of f(x)=-(6x^2)/((x^3-1)^2) is -(12x(-2x^3-1))/((x^3-1)^3)

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

The first derivative of f(x)=-(6x^2)/((x^3-1)^2) is -(12x(-2x^3-1))/((x^3-1)^3)

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derivative of

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

- \frac{12 x ( - 2 x ^{3} - 1 )}{( x ^{3} - 1 ) ^{3}}

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

- 6 \cdot \frac{2 x ( x ^{3} - 1 ) ^{2} - 6 x ^{2} ( x ^{3} - 1 ) x ^{2}}{( ( x ^{3} - 1 ) ^{2} ) ^{2}} : - \frac{12 x ( - 2 x ^{3} - 1 )}{( x ^{3} - 1 ) ^{3}}

= - \frac{12 x ( - 2 x ^{3} - 1 )}{( x ^{3} - 1 ) ^{3}}

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

\frac{d}{d x} (sin (k x))

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

\int 2 u^{2} d u

x \to - 1 lim (\frac{x - 1}{x + 1})

What is the derivative of f(x)=-(6x^2)/((x^3-1)^2) ?
The derivative of f(x)=-(6x^2)/((x^3-1)^2) is -(12x(-2x^3-1))/((x^3-1)^3)
What is the first derivative of f(x)=-(6x^2)/((x^3-1)^2) ?
The first derivative of f(x)=-(6x^2)/((x^3-1)^2) is -(12x(-2x^3-1))/((x^3-1)^3)