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

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

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

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

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

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

Solution steps

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

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

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

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

Simplify

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

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

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

\int x 5^{x^{2}} d x

\frac{d}{d x} (2 x^{3} - 3 x^{2} - 5)

f (x) = 5 cos^{2} (x)

x \to 2 lim (2 - x)

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