What is the derivative of x/(3x-4*ln(x)) ?

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

What is the first derivative of x/(3x-4*ln(x)) ?

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

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\frac{d}{d x} (\frac{x}{3 x - 4 \cdot ln ( x )})

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

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d x}{d x} = 1

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

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

Simplify

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

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

f (x) = x^{2} 16 - x^{2}

\frac{\partial}{\partial x} (x \cdot e^{x^{2} y})

f (x) = 3 x^{2} cos (x) + 7 x

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

\frac{d}{d x} (8 - x^{2})

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