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

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

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

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

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derivative of

y = \frac{x ^{3} - x}{x + 1}

2 x - 1

Solution steps

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

Simplify

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

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = x - 1

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

\frac{d x}{d x} = 1

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

= 1 \cdot (x - 1) + 1 \cdot x

Simplify

1 \cdot (x - 1) + 1 \cdot x : 2 x - 1

= 2 x - 1

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