# Find the derivative of f (x) = {1 + log square root x} / x.

## Question:

Find the derivative of

{eq}\displaystyle f (x) = \dfrac {1 + \log \sqrt x} x {/eq}.

## Quotient Rule for Derivatives:

If a function is given in the form {eq}f(x) = \dfrac{u}{v} {/eq},

then the derivative of the given function can be obtained by using the quotient rule for derivatives:

{eq}f'(x) = \dfrac{u'v - uv'}{v^2} {/eq}

Here, {eq}u' {/eq} and {eq}v' {/eq} are the derivatives of {eq}u {/eq} and {eq}v {/eq} with respect to {eq}x {/eq}.

The given function is:

{eq}f(x) = \dfrac{1 + \log(\sqrt{x})}{x} {/eq}

Using:

{eq}\log(a) = \dfrac{\ln(a)}{\ln(10)} {/eq}

We get:

{eq}f(x) = \dfrac{1 + \dfrac{\ln(\sqrt{x})}{\ln(10)}}{x} \\ f(x) = \dfrac{\ln(10) + \ln(\sqrt{x})}{x\ln(10)} {/eq}

Using:

{eq}\ln(a^b) = b \ln(a) {/eq}

We get:

{eq}f(x) = \dfrac{\ln(10) + \dfrac{1}{2}\ln(x)}{x\ln(10)} \\ f(x) = \dfrac{2\ln(10) + \ln(x)}{2x\ln(10)} {/eq}

We have:

{eq}u = 2\ln(10) + \ln(x), \quad v = 2x\ln(10) {/eq}

Differentiating u and v with respect to x, we get:

{eq}u' = 0 + \dfrac{1}{x}, \quad v' = 2(1)\ln(10) \\ u' = \dfrac{1}{x}, \quad v' = 2\ln(10) {/eq}

Finding the derivative of f(x) by using the quotient rule of derivative:

{eq}f'(x) = \dfrac{u'v - uv'}{v^2} \\ f'(x) = \dfrac{\left(\dfrac{1}{x}\right)2x\ln(10) - (2\ln(10) + \ln(x))2\ln(10)}{(2x\ln(10))^2} \\ f'(x) = \dfrac{2\ln(10) - 2\ln(10)(2\ln(10) + \ln(x))}{2\ln(10)(2x^2\ln(10))} \\ f'(x) = \dfrac{2\ln(10)(1 - 2\ln(10) - \ln(x))}{2\ln(10)(2x^2\ln(10))} \\ \boxed{f'(x) = \dfrac{1 - 2\ln(10) - \ln(x)}{2x^2\ln(10)}} {/eq}

This is the derivative of the given function f(x). 