What is the derivative of (2x+4x^2/((1+4x)^2)) ?

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

x \to - 5 - lim (\frac{x}{x + 5})

(50, 43), (60, 86)

\frac{d y}{d x} = - 10 x y^{2}

\int_{0}^{\infty} e^{- 8 x} d x

y^{^{''}} - 6 y^{^{'}} + 9 y = t^{- 5} e^{3 t}

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