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

The derivative of y=(2x)/(x^2+1) is (2(-x^2+1))/((x^2+1)^2)

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

The first derivative of y=(2x)/(x^2+1) is (2(-x^2+1))/((x^2+1)^2)

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derivative of

y = \frac{2 x}{x ^{2} + 1}

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

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) = 2 x

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

Simplify

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

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

\int t^{2} (t - \frac{2}{t}) d t

y^{^{'}} = \frac{( 4 x sec ( \frac{y}{x} ) + y )}{x}

x \to - 6 lim (\frac{6}{x + 6})

y^{^{''}} + y = 8 x sin (x)

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

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