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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

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

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

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

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

Simplify

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

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

\int (6 x^{3} + 3 x^{2}) d x

d er i v a t i v e ln (sec (x))

x \to 0 lim (x^{2} cos (\frac{6}{x}))

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

\frac{\partial}{\partial s} (s + 2 t)

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