What is the derivative of y= x/(1-ln(x-5)) ?

The derivative of y= x/(1-ln(x-5)) is (2x-ln(x-5)(x-5)-5)/((x-5)(1-ln(x-5))^2)

What is the first derivative of y= x/(1-ln(x-5)) ?

The first derivative of y= x/(1-ln(x-5)) is (2x-ln(x-5)(x-5)-5)/((x-5)(1-ln(x-5))^2)

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derivative of

y = \frac{x}{1 - ln ( x - 5 )}

\frac{2 x - ln ( x - 5 ) ( x - 5 ) - 5}{( x - 5 ) ( 1 - ln ( x - 5 ) ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x}{1 - ln ( x - 5 )})

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 - 5 ) ) - \frac{d}{d x} ( 1 - ln ( x - 5 ) ) x}{( 1 - ln ( x - 5 ) ) ^{2}}

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

\frac{d}{d x} (1 - ln (x - 5)) = - \frac{1}{x - 5}

= \frac{1 \cdot ( 1 - ln ( x - 5 ) ) - ( - \frac{1}{x - 5} ) x}{( 1 - ln ( x - 5 ) ) ^{2}}

Simplify

\frac{1 \cdot ( 1 - ln ( x - 5 ) ) - ( - \frac{1}{x - 5} ) x}{( 1 - ln ( x - 5 ) ) ^{2}} : \frac{2 x - ln ( x - 5 ) ( x - 5 ) - 5}{( x - 5 ) ( 1 - ln ( x - 5 ) ) ^{2}}

= \frac{2 x - ln ( x - 5 ) ( x - 5 ) - 5}{( x - 5 ) ( 1 - ln ( x - 5 ) ) ^{2}}

What is the derivative of y= x/(1-ln(x-5)) ?
The derivative of y= x/(1-ln(x-5)) is (2x-ln(x-5)(x-5)-5)/((x-5)(1-ln(x-5))^2)
What is the first derivative of y= x/(1-ln(x-5)) ?
The first derivative of y= x/(1-ln(x-5)) is (2x-ln(x-5)(x-5)-5)/((x-5)(1-ln(x-5))^2)