What is the derivative of (4+3x/(x^2-2x)) ?

The derivative of (4+3x/(x^2-2x)) is (-3x^2-8x+8)/((x^2-2x)^2)

What is the first derivative of (4+3x/(x^2-2x)) ?

The first derivative of (4+3x/(x^2-2x)) is (-3x^2-8x+8)/((x^2-2x)^2)

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\frac{d}{d x} (\frac{4 + 3 x}{x ^{2} - 2 x})

\frac{- 3 x ^{2} - 8 x + 8}{( x ^{2} - 2 x ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{4 + 3 x}{x ^{2} - 2 x})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 4 + 3 x ) ( x ^{2} - 2 x ) - \frac{d}{d x} ( x ^{2} - 2 x ) ( 4 + 3 x )}{( x ^{2} - 2 x ) ^{2}}

\frac{d}{d x} (4 + 3 x) = 3

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

= \frac{3 ( x ^{2} - 2 x ) - ( 2 x - 2 ) ( 4 + 3 x )}{( x ^{2} - 2 x ) ^{2}}

Expand

3 (x^{2} - 2 x) - (2 x - 2) (4 + 3 x) : - 3 x^{2} - 8 x + 8

= \frac{- 3 x ^{2} - 8 x + 8}{( x ^{2} - 2 x ) ^{2}}

What is the derivative of (4+3x/(x^2-2x)) ?
The derivative of (4+3x/(x^2-2x)) is (-3x^2-8x+8)/((x^2-2x)^2)
What is the first derivative of (4+3x/(x^2-2x)) ?
The first derivative of (4+3x/(x^2-2x)) is (-3x^2-8x+8)/((x^2-2x)^2)