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

The derivative of (x^2/((3+x))) is (6x+x^2)/((3+x)^2)

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

The first derivative of (x^2/((3+x))) is (6x+x^2)/((3+x)^2)

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

\frac{6 x + x ^{2}}{( 3 + x ) ^{2}}

Solution steps

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

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

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

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

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

Simplify

\frac{2 x ( 3 + x ) - 1 \cdot x ^{2}}{( 3 + x ) ^{2}} : \frac{6 x + x ^{2}}{( 3 + x ) ^{2}}

= \frac{6 x + x ^{2}}{( 3 + x ) ^{2}}

\frac{\partial}{\partial x} (x^{7})

\frac{d}{d y} ((\frac{1}{y ^{2}} - \frac{3}{y ^{4}}) (y + 9 y^{3}))

2 csc (1 - 3 x)

f (x) = (2 x + 3)^{5}, at x = - 1

What is the derivative of (x^2/((3+x))) ?
The derivative of (x^2/((3+x))) is (6x+x^2)/((3+x)^2)
What is the first derivative of (x^2/((3+x))) ?
The first derivative of (x^2/((3+x))) is (6x+x^2)/((3+x)^2)