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

The derivative of f(x)= 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 f(x)= x/(1-ln(x-4)) ?

The first derivative of f(x)= 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

f (x) = \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) = (1 + 3 x) (5 - 4 x), at x = 1

\frac{d}{d x} (a cos (a x))

\int \frac{x ^{3} + 4}{x ^{2} + 4} d x

\frac{\partial}{\partial x} (cos (x - y) + (x + y))

y = x, y = 2 x 25 - x^{2}

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