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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{d N}{d t} + N = Nt e^{t + 2}

\int \frac{x ^{4}}{x ^{3} - 1} d x

y = \frac{4}{4 - x ^{2}}, (0, 1)

t y^{^{'}} + y = 8

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