What is the (x/(4x-3ln(x)))^' ?

The (x/(4x-3ln(x)))^' is (-3ln(x)+3)/((4x-3ln(x))^2)

What is the first (x/(4x-3ln(x)))^' ?

The first (x/(4x-3ln(x)))^' is (-3ln(x)+3)/((4x-3ln(x))^2)

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(\frac{x}{4 x - 3 ln ( x )})^{^{'}}

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

Solution steps

(\frac{x}{4 x - 3 ln ( x )})^{'}

Apply the Quotient Rule:

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

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

x^{'} = 1

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

= \frac{1 \cdot ( 4 x - 3 ln ( x ) ) - ( 4 - \frac{3}{x} ) x}{( 4 x - 3 ln ( x ) ) ^{2}}

Simplify

\frac{1 \cdot ( 4 x - 3 ln ( x ) ) - ( 4 - \frac{3}{x} ) x}{( 4 x - 3 ln ( x ) ) ^{2}} : \frac{- 3 ln ( x ) + 3}{( 4 x - 3 ln ( x ) ) ^{2}}

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

\frac{5 s + 31}{s ^{2} + 6 s + 9}

\int_{1}^{3} x^{- 2} d x

x \to 2 + lim (\frac{5}{x - 2})

x \to \infty lim (- 2)

\int y cos (x y) - 3 y^{2} + e^{x} d x

What is the (x/(4x-3ln(x)))^' ?
The (x/(4x-3ln(x)))^' is (-3ln(x)+3)/((4x-3ln(x))^2)
What is the first (x/(4x-3ln(x)))^' ?
The first (x/(4x-3ln(x)))^' is (-3ln(x)+3)/((4x-3ln(x))^2)