What is the derivative of (x^2/((4+x))) ?

The derivative of (x^2/((4+x))) is (8x+x^2)/((4+x)^2)

What is the first derivative of (x^2/((4+x))) ?

The first derivative of (x^2/((4+x))) is (8x+x^2)/((4+x)^2)

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

\frac{8 x + x ^{2}}{( 4 + x ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

\frac{d}{d x} (4 + x) = 1

= \frac{2 x ( 4 + x ) - 1 \cdot x ^{2}}{( 4 + x ) ^{2}}

Simplify

\frac{2 x ( 4 + x ) - 1 \cdot x ^{2}}{( 4 + x ) ^{2}} : \frac{8 x + x ^{2}}{( 4 + x ) ^{2}}

= \frac{8 x + x ^{2}}{( 4 + x ) ^{2}}

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

y^{2} = x^{2} (x + 1), (3, 6)

\int x^{l n (x)} d x

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

1 - \frac{1}{x}

What is the derivative of (x^2/((4+x))) ?
The derivative of (x^2/((4+x))) is (8x+x^2)/((4+x)^2)
What is the first derivative of (x^2/((4+x))) ?
The first derivative of (x^2/((4+x))) is (8x+x^2)/((4+x)^2)