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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

y^{^{'}} = 4 y + 1

\int_{2}^{4} \frac{6 - x - y}{8} d y

n = 1 \sum \infty \frac{n}{2 ^{n} + 5}

\frac{\partial}{\partial y} (- e^{x} cos (y))

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