# Find the derivative or the given function f(x)=\frac{x-5}{\ln x}

## Question:

Find the derivative or the given function {eq}f(x)=\frac{x-5}{\ln x} {/eq}

## Differentiation

If we have a function with one independent variable, we can find its derivative if we know differentiation rules, for example, if the function is a quotient function its derivative is: {eq}\displaystyle \ f(x) \, = \,\frac{ a }{ b } \; \Rightarrow \; \ f'(x) \; = \; \frac{ a' \times b \; - \; a \times b' }{ b^2 } {/eq}

The function is:

{eq}\displaystyle \ f(x) \, = \, \frac{x-5}{\ln x} {/eq}

It is a quotient function, and its derivative is:

{eq}\displaystyle \ f'(x) \; = \; \frac{ \frac{d}{dx}\left[ x-5 \right] \; \times \; \ln(x) \; - \; \left( (x-5) \times \frac{d}{dx}\left[ \ln(x) \right] \right) }{ \left( \ln(x) \right)^2 } \\ \displaystyle \ f'(x) \; = \; \frac{ (1)(\ln(x)) \; - \; \left( (x-5) \times \frac{ 1 }{ x } \right) }{ (\ln(x))^2 }\\ \displaystyle \ f'(x) \; = \; \frac{ \ln(x) \; - \; (1-5/x) }{ \ln^2(x) } \\ \displaystyle \ f'(x) \; = \; \frac{ 1 }{\ln(x) } \; - \; \frac{ 1 }{ \ln^2(x) } \; + \; \frac{ 5 }{ x^2\ln^2(x) } \\ \; \text{ or } \; \\ \displaystyle \ f'(x) \; = \; - \left( \ln \left( x \right) \right) ^{-1}-{\frac {5-x}{ \left( \ln \left( x \right) \right) ^{2}x}} \\ {/eq}