What is the derivative of (x-1/(x^2+2x+6)) ?

The derivative of (x-1/(x^2+2x+6)) is (-x^2+2x+8)/((x^2+2x+6)^2)

What is the first derivative of (x-1/(x^2+2x+6)) ?

The first derivative of (x-1/(x^2+2x+6)) is (-x^2+2x+8)/((x^2+2x+6)^2)

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\frac{d}{d x} (\frac{x - 1}{x ^{2} + 2 x + 6})

\frac{- x ^{2} + 2 x + 8}{( x ^{2} + 2 x + 6 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x - 1}{x ^{2} + 2 x + 6})

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} + 2 x + 6 ) - \frac{d}{d x} ( x ^{2} + 2 x + 6 ) ( x - 1 )}{( x ^{2} + 2 x + 6 ) ^{2}}

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

\frac{d}{d x} (x^{2} + 2 x + 6) = 2 x + 2

= \frac{1 \cdot ( x ^{2} + 2 x + 6 ) - ( 2 x + 2 ) ( x - 1 )}{( x ^{2} + 2 x + 6 ) ^{2}}

Simplify

\frac{1 \cdot ( x ^{2} + 2 x + 6 ) - ( 2 x + 2 ) ( x - 1 )}{( x ^{2} + 2 x + 6 ) ^{2}} : \frac{- x ^{2} + 2 x + 8}{( x ^{2} + 2 x + 6 ) ^{2}}

= \frac{- x ^{2} + 2 x + 8}{( x ^{2} + 2 x + 6 ) ^{2}}

y^{^{'}} - 2 y = - 5 x^{2}

f (x) = 3 e^{- 4 x}, at x = 0

\frac{s}{s ^{2} - 2 s + 10}

\frac{d}{d x} (\frac{6 x ^{2} + 4 x + 2}{x})

\frac{d}{d y} (\frac{y ^{4} + 48}{24 y})

What is the derivative of (x-1/(x^2+2x+6)) ?
The derivative of (x-1/(x^2+2x+6)) is (-x^2+2x+8)/((x^2+2x+6)^2)
What is the first derivative of (x-1/(x^2+2x+6)) ?
The first derivative of (x-1/(x^2+2x+6)) is (-x^2+2x+8)/((x^2+2x+6)^2)