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

The derivative of (x^2+4x)/((2+x)^2) is 8/((2+x)^3)

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

The first derivative of (x^2+4x)/((2+x)^2) is 8/((2+x)^3)

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derivative of

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

\frac{8}{( 2 + x ) ^{3}}

Solution steps

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

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

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

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

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

Simplify

\frac{( 2 x + 4 ) ( 2 + x ) ^{2} - 2 ( 2 + x ) ( x ^{2} + 4 x )}{( ( 2 + x ) ^{2} ) ^{2}} : \frac{8}{( 2 + x ) ^{3}}

= \frac{8}{( 2 + x ) ^{3}}

ln (1 - 2 x)

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

\frac{\partial}{\partial y} (x^{6 y})

y^{^{''}} + y^{^{'}} - 6 y = 0, y (0) = 1, y^{^{'}} (0) = k

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

What is the derivative of (x^2+4x)/((2+x)^2) ?
The derivative of (x^2+4x)/((2+x)^2) is 8/((2+x)^3)
What is the first derivative of (x^2+4x)/((2+x)^2) ?
The first derivative of (x^2+4x)/((2+x)^2) is 8/((2+x)^3)