What is the derivative of x/(1-ln(x-4)) ?

The derivative of x/(1-ln(x-4)) is (2x-ln(x-4)(x-4)-4)/((x-4)(1-ln(x-4))^2)

What is the first derivative of x/(1-ln(x-4)) ?

The first derivative of x/(1-ln(x-4)) is (2x-ln(x-4)(x-4)-4)/((x-4)(1-ln(x-4))^2)

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derivative of

\frac{x}{1 - ln ( x - 4 )}

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

Solution steps

\frac{d}{d x} (\frac{x}{1 - ln ( x - 4 )})

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (1 - ln (x - 4)) = - \frac{1}{x - 4}

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

Simplify

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

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

f (x) = 2 x^{2} - 3

x - sin (x)

\frac{\partial}{\partial x} (sin (\frac{2 y}{x}))

\frac{d}{d x} (\frac{6}{( 1 + x ) ^{4}})

\frac{d}{d x} (\frac{6 x ^{3} - 2}{x ^{2} - 4})

What is the derivative of x/(1-ln(x-4)) ?
The derivative of x/(1-ln(x-4)) is (2x-ln(x-4)(x-4)-4)/((x-4)(1-ln(x-4))^2)
What is the first derivative of x/(1-ln(x-4)) ?
The first derivative of x/(1-ln(x-4)) is (2x-ln(x-4)(x-4)-4)/((x-4)(1-ln(x-4))^2)