What is the derivative of f(x)=(4x^3-4)/(x^3-8) ?

The derivative of f(x)=(4x^3-4)/(x^3-8) is -(84x^2)/((x^3-8)^2)

What is the first derivative of f(x)=(4x^3-4)/(x^3-8) ?

The first derivative of f(x)=(4x^3-4)/(x^3-8) is -(84x^2)/((x^3-8)^2)

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derivative of

f (x) = \frac{4 x ^{3} - 4}{x ^{3} - 8}

- \frac{84 x ^{2}}{( x ^{3} - 8 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{4 x ^{3} - 4}{x ^{3} - 8})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 4 x ^{3} - 4 ) ( x ^{3} - 8 ) - \frac{d}{d x} ( x ^{3} - 8 ) ( 4 x ^{3} - 4 )}{( x ^{3} - 8 ) ^{2}}

\frac{d}{d x} (4 x^{3} - 4) = 12 x^{2}

\frac{d}{d x} (x^{3} - 8) = 3 x^{2}

= \frac{12 x ^{2} ( x ^{3} - 8 ) - 3 x ^{2} ( 4 x ^{3} - 4 )}{( x ^{3} - 8 ) ^{2}}

Simplify

\frac{12 x ^{2} ( x ^{3} - 8 ) - 3 x ^{2} ( 4 x ^{3} - 4 )}{( x ^{3} - 8 ) ^{2}} : - \frac{84 x ^{2}}{( x ^{3} - 8 ) ^{2}}

= - \frac{84 x ^{2}}{( x ^{3} - 8 ) ^{2}}

\frac{1}{x} \frac{d y}{d x} - ln (x) cot (y) = 0

\frac{d}{d x} ((x + 5) cos (x))

\int_{0}^{3} (\frac{1}{2} x - 1) d x

y ln (y) d x + x d y = 0

y = x^{3} - 3 x^{2} - 9 x + 27, at x = 4

What is the derivative of f(x)=(4x^3-4)/(x^3-8) ?
The derivative of f(x)=(4x^3-4)/(x^3-8) is -(84x^2)/((x^3-8)^2)
What is the first derivative of f(x)=(4x^3-4)/(x^3-8) ?
The first derivative of f(x)=(4x^3-4)/(x^3-8) is -(84x^2)/((x^3-8)^2)