What is the derivative of (x^3-4/(x-1)) ?

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

What is the first derivative of (x^3-4/(x-1)) ?

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{d y}{d x} = \frac{ln ( x )}{x y}, y (1) = 2

y = \frac{( t ^{3} + t )}{( t ^{4} - 2 )}

\int \frac{1}{( y ^{2} - 1 )} d y

cos^{3} (x) + sin^{4} (x)

\int \frac{3 x - 2}{x + 1} d x

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