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

The derivative of g(x)=(4x^2-x+2)/(3-4x) is (-16x^2+24x+5)/((3-4x)^2)

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

The first derivative of g(x)=(4x^2-x+2)/(3-4x) is (-16x^2+24x+5)/((3-4x)^2)

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derivative of

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

\frac{- 16 x ^{2} + 24 x + 5}{( 3 - 4 x ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (4 x^{2} - x + 2) = 8 x - 1

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

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

Simplify

\frac{( 8 x - 1 ) ( 3 - 4 x ) - ( - 4 ) ( 4 x ^{2} - x + 2 )}{( 3 - 4 x ) ^{2}} : \frac{- 16 x ^{2} + 24 x + 5}{( 3 - 4 x ) ^{2}}

= \frac{- 16 x ^{2} + 24 x + 5}{( 3 - 4 x ) ^{2}}

y = 3 x^{2} + 2 xx a

x \to 0 lim (x^{6} (cos (\frac{9}{x})))

y = e^{- 8 x}, x \geq 0

\frac{d}{d x} (sin (7 x) + sin^{2} (7 x))

n = 1 \sum \infty \frac{1}{n ^{2} + 2 n + 2}

What is the derivative of g(x)=(4x^2-x+2)/(3-4x) ?
The derivative of g(x)=(4x^2-x+2)/(3-4x) is (-16x^2+24x+5)/((3-4x)^2)
What is the first derivative of g(x)=(4x^2-x+2)/(3-4x) ?
The first derivative of g(x)=(4x^2-x+2)/(3-4x) is (-16x^2+24x+5)/((3-4x)^2)