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

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

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

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

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

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

Solution steps

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

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

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

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

Simplify

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

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

x \to 0 lim (x^{3 s i n (x)})

\int (7 x^{3} - 2 x^{2} + 4) d x

\int 9 tan^{2} (x) d x

\int_{- 1}^{2} \frac{1}{x ^{4}} d x

y^{^{''}} + 2 y^{^{'}} + y = 0, y (1) = 1, y^{^{'}} (1) = 0

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