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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derivative of

\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}}

\frac{d}{d x} (e^{k^{2} x})

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

(x^{2} + 1) y^{^{''}} + 2 x y^{^{'}} = 0

y d x - (y - x) d y = 0

y^{^{'}} + \frac{1}{2} y = 0

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)