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

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

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

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

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

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

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

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

\frac{d x}{d x} = 1

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

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

Simplify

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

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

x \to 2 lim (a^{2} x + 2 a)

(1 + y) x d x - (1 + x) y d y = 0

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

\frac{d}{d x} (e^{- \frac{1}{2} x})

\frac{3 e ^{- 7 s}}{s ^{2}} + \frac{7 e ^{- 7 s}}{s} + \frac{2 e ^{- 7 s}}{s}

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