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

The derivative of (2x^2-4/((x+4)^2)) is (8(2x+1))/((x+4)^3)

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

The first derivative of (2x^2-4/((x+4)^2)) is (8(2x+1))/((x+4)^3)

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

\frac{8 ( 2 x + 1 )}{( x + 4 ) ^{3}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{4 x ( x + 4 ) ^{2} - 2 ( x + 4 ) ( 2 x ^{2} - 4 )}{( ( x + 4 ) ^{2} ) ^{2}} : \frac{8 ( 2 x + 1 )}{( x + 4 ) ^{3}}

= \frac{8 ( 2 x + 1 )}{( x + 4 ) ^{3}}

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 2 e^{- x} + 3

f (x) = \frac{1}{4 ( 1 + x ^{2} )}, g (x) = \frac{1}{8} x^{2}

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

x \to 45 lim (\frac{x ^{2}}{9} - \frac{5}{x})

\frac{d}{d x} (sin^{5} (2 x))

What is the derivative of (2x^2-4/((x+4)^2)) ?
The derivative of (2x^2-4/((x+4)^2)) is (8(2x+1))/((x+4)^3)
What is the first derivative of (2x^2-4/((x+4)^2)) ?
The first derivative of (2x^2-4/((x+4)^2)) is (8(2x+1))/((x+4)^3)