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

The derivative of (4x-3/(x^2-2x)) is (-4x^2+6x-6)/((x^2-2x)^2)

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

The first derivative of (4x-3/(x^2-2x)) is (-4x^2+6x-6)/((x^2-2x)^2)

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

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

Solution steps

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

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

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

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

Expand

4 (x^{2} - 2 x) - (2 x - 2) (4 x - 3) : - 4 x^{2} + 6 x - 6

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

f (x) = tan^{7} (x)

e^{x} + \frac{1}{1 + x}, 0

\int 19 tan^{3} (θ) d θ

\frac{\partial}{\partial y} (x^{5} - 6 x + 4 y^{3} - y^{2})

What is the derivative of (4x-3/(x^2-2x)) ?
The derivative of (4x-3/(x^2-2x)) is (-4x^2+6x-6)/((x^2-2x)^2)
What is the first derivative of (4x-3/(x^2-2x)) ?
The first derivative of (4x-3/(x^2-2x)) is (-4x^2+6x-6)/((x^2-2x)^2)