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

The derivative of (x-4/(x^2+1)) is (-x^2+8x+1)/((x^2+1)^2)

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

The first derivative of (x-4/(x^2+1)) is (-x^2+8x+1)/((x^2+1)^2)

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

\frac{- x ^{2} + 8 x + 1}{( x ^{2} + 1 ) ^{2}}

Solution steps

\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}}

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

\frac{d}{d x} (x - 4) = 1

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

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

Simplify

\frac{1 \cdot ( x ^{2} + 1 ) - 2 x ( x - 4 )}{( x ^{2} + 1 ) ^{2}} : \frac{- x ^{2} + 8 x + 1}{( x ^{2} + 1 ) ^{2}}

= \frac{- x ^{2} + 8 x + 1}{( x ^{2} + 1 ) ^{2}}

n = 0 \sum \infty (- \frac{1}{2})^{n + 1}

\frac{d}{d x} (I n (3 x))

\int x^{- \frac{1}{2}} e^{- x} d x

\frac{\partial}{\partial y} (x^{0.5} y^{0.5})

\frac{d}{d x} (cos (3 x^{2} - 2 x))

What is the derivative of (x-4/(x^2+1)) ?
The derivative of (x-4/(x^2+1)) is (-x^2+8x+1)/((x^2+1)^2)
What is the first derivative of (x-4/(x^2+1)) ?
The first derivative of (x-4/(x^2+1)) is (-x^2+8x+1)/((x^2+1)^2)