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

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

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

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

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

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

Solution steps

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

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

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

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

Simplify

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

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

\frac{d}{d x} (\frac{- x ^{2} + 64}{( x ^{2} + 64 ) ^{2}})

f (x) = cos (x) + x sin (x)

\frac{\partial}{\partial x} (4 x + 9 y)

\frac{\partial}{\partial x} (x^{4} + 5 x y^{9})

t \to 0 lim (t^{2})

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