What is the derivative of (x-2/(x^2+5x+4)) ?

The derivative of (x-2/(x^2+5x+4)) is (-x^2+4x+14)/((x^2+5x+4)^2)

What is the first derivative of (x-2/(x^2+5x+4)) ?

The first derivative of (x-2/(x^2+5x+4)) is (-x^2+4x+14)/((x^2+5x+4)^2)

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\frac{d}{d x} (\frac{x - 2}{x ^{2} + 5 x + 4})

\frac{- x ^{2} + 4 x + 14}{( x ^{2} + 5 x + 4 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x - 2}{x ^{2} + 5 x + 4})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x - 2 ) ( x ^{2} + 5 x + 4 ) - \frac{d}{d x} ( x ^{2} + 5 x + 4 ) ( x - 2 )}{( x ^{2} + 5 x + 4 ) ^{2}}

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

\frac{d}{d x} (x^{2} + 5 x + 4) = 2 x + 5

= \frac{1 \cdot ( x ^{2} + 5 x + 4 ) - ( 2 x + 5 ) ( x - 2 )}{( x ^{2} + 5 x + 4 ) ^{2}}

Simplify

\frac{1 \cdot ( x ^{2} + 5 x + 4 ) - ( 2 x + 5 ) ( x - 2 )}{( x ^{2} + 5 x + 4 ) ^{2}} : \frac{- x ^{2} + 4 x + 14}{( x ^{2} + 5 x + 4 ) ^{2}}

= \frac{- x ^{2} + 4 x + 14}{( x ^{2} + 5 x + 4 ) ^{2}}

\frac{d y}{d x} = - y^{\frac{1}{3}}

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

y = 2 x^{3}, y = 18 x

f (x) = 3 x - 2 x^{2}, (- 3, - 27)

x \to 0 lim (π^{2})

What is the derivative of (x-2/(x^2+5x+4)) ?
The derivative of (x-2/(x^2+5x+4)) is (-x^2+4x+14)/((x^2+5x+4)^2)
What is the first derivative of (x-2/(x^2+5x+4)) ?
The first derivative of (x-2/(x^2+5x+4)) is (-x^2+4x+14)/((x^2+5x+4)^2)