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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

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

x^{3} y^{^{'''}} - 6 y = 0

\int 7 arctan (3 y) d y

\int \frac{6 sin ( 2 x )}{sin ( x )} d x

n = 1 \sum \infty \frac{8}{π ^{n}}

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