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

The derivative of (2x^2-4x+3/(2-3x)) is (-6x^2+8x+1)/((2-3x)^2)

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

The first derivative of (2x^2-4x+3/(2-3x)) is (-6x^2+8x+1)/((2-3x)^2)

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

\frac{- 6 x ^{2} + 8 x + 1}{( 2 - 3 x ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{( 4 x - 4 ) ( 2 - 3 x ) - ( - 3 ) ( 2 x ^{2} - 4 x + 3 )}{( 2 - 3 x ) ^{2}} : \frac{- 6 x ^{2} + 8 x + 1}{( 2 - 3 x ) ^{2}}

= \frac{- 6 x ^{2} + 8 x + 1}{( 2 - 3 x ) ^{2}}

\int f (r) d r

\int \frac{t}{t ^{4} + 6} d t

\int \frac{x ^{4}}{3 - x ^{5}} d x

\frac{d}{d x} (\frac{a x + b}{a x - b})

f (x) = x^{2} + 4, (3, 13)

What is the derivative of (2x^2-4x+3/(2-3x)) ?
The derivative of (2x^2-4x+3/(2-3x)) is (-6x^2+8x+1)/((2-3x)^2)
What is the first derivative of (2x^2-4x+3/(2-3x)) ?
The first derivative of (2x^2-4x+3/(2-3x)) is (-6x^2+8x+1)/((2-3x)^2)