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

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

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

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

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

1

Solution steps

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

Simplify

\frac{x ^{2} + x}{x + 1} : x

= \frac{d x}{d x}

Apply the common derivative:

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

= 1

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

y^{^{''}} - 4 y^{^{'}} + 13 y = 0, y (0) = 1, y^{^{'}} (0) = 4

\int x^{2} + 1 x d x

x \to 0 lim (sin (x) \cdot ln (4 x))

\frac{\partial}{\partial x} (2 x + y - 5)

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