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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\int \frac{cos ( x )}{1 + cos ( x )} d x

\frac{d y}{d x} = \frac{1 - x ^{2}}{y ^{2}}

t + t^{3}

\frac{d y}{d x} = 8 - 4 y

\int \frac{e ^{2 x}}{3 x} d x

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