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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

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

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

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

Simplify

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

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

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

(2 x y + y^{2}) d x - 2 x^{2} d y = 0

\frac{d}{d x} (\frac{( x - 4 )}{x} + ln (\frac{x}{4}))

\frac{\partial}{\partial y} (\frac{x}{2 y ^{4}})

3 x + 1

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