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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

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

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

Solution steps

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

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

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

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

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

Simplify

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

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

t \to 1 lim ((t^{3} + 3)^{4})

\frac{\partial}{\partial x} (e^{x^{7}})

\frac{\partial}{\partial x} (e^{x^{3}})

y^{^{'}} - \frac{1}{2} y = (5 x - 5) y^{- 1}, y (0) = 4

\int \frac{x ^{5}}{1 - x ^{4}} d x

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