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}}

\int \frac{x ^{6} - x ^{3} + 16}{x ^{4} + 4 x ^{2}} d x

t \frac{d y}{d t} = t^{3} + 15 t^{3} y

- 6 y^{^{'}} + 18 x^{2} = 0

(t^{3}) d x + (x t^{2} + x^{3}) d t = 0

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)