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

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

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

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

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

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

Solution steps

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

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

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

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

Simplify

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

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

f (x) = - x^{\frac{4}{3}}

\int_{- 1}^{1} (8 - 8 x^{2})^{2} d x

\frac{\partial}{\partial x} (y^{2} cos (x y))

\int a cos (x) d x

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

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