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

The derivative of (x^2-4x+3/(2x^2-3x+2)) is (5x^2-8x+1)/((2x^2-3x+2)^2)

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

The first derivative of (x^2-4x+3/(2x^2-3x+2)) is (5x^2-8x+1)/((2x^2-3x+2)^2)

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

\frac{5 x ^{2} - 8 x + 1}{( 2 x ^{2} - 3 x + 2 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Expand

(2 x - 4) (2 x^{2} - 3 x + 2) - (4 x - 3) (x^{2} - 4 x + 3) : 5 x^{2} - 8 x + 1

= \frac{5 x ^{2} - 8 x + 1}{( 2 x ^{2} - 3 x + 2 ) ^{2}}

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

\frac{\partial}{\partial x} (2 x^{2} + y^{2} + 2 x y + y x + 7 y)

\frac{\partial}{\partial v} (ln (u cos (v)))

\frac{\partial}{\partial y} (ln (3 x y) + x^{2})

\frac{\partial}{\partial y} (\frac{y}{( 2 x + 3 y )})

What is the derivative of (x^2-4x+3/(2x^2-3x+2)) ?
The derivative of (x^2-4x+3/(2x^2-3x+2)) is (5x^2-8x+1)/((2x^2-3x+2)^2)
What is the first derivative of (x^2-4x+3/(2x^2-3x+2)) ?
The first derivative of (x^2-4x+3/(2x^2-3x+2)) is (5x^2-8x+1)/((2x^2-3x+2)^2)