# Differentiate the given function. Give your answer in terms of natural logs with the arguments in...

## Question:

Differentiate the given function. Give your answer in terms of natural logs with the arguments in parentheses. {eq}f(x) = \frac{ \log_7 x}{9 \sqrt x} {/eq}

## Quotient Rule:

Let {eq}a(x) {/eq} and {eq}b(x) {/eq} are the function of {eq}x {/eq} then the derivative of {eq}\frac{{a(x)}}{{b(x)}} {/eq} is given by:

{eq}{\,\frac{d}{{dx}}\frac{a}{b} = \frac{{b\frac{{da}}{{dx}} - a\frac{{db}}{{dx}}}}{{{b^2}}}} {/eq}

The following rules are relevant to this problem:

1.{eq}{{{\log }_7}a = \frac{{\ln a}}{{\ln 7}}} {/eq}

2.{eq}{\frac{d}{{dx}}cw\left( x \right) = c\frac{d}{{dx}}w\left( x \right)} {/eq}

3.{eq}{\frac{d}{{dx}}\left( {\ln x} \right) = \frac{1}{x},\frac{d}{{dx}}{x^n} = n{x^{n - 1}}} {/eq}

## Answer and Explanation:

Given that: {eq}\displaystyle f(x) = \frac{{{{\log }_7}x}}{{9\sqrt x }} {/eq}

{eq}\displaystyle \eqalign{ & f(x) = \frac{{{{\log }_7}x}}{{9\sqrt x }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\left( {\frac{{\ln x}}{{\ln 7}}} \right)}}{{9\sqrt x }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{{\log }_7}a = \frac{{\ln a}}{{\ln 7}}} \right) \cr & f(x) = \left( {\frac{1}{{9\ln 7}}} \right)\frac{{\ln x}}{{\sqrt x }} \cr & {\text{Differentiating }}f(x){\text{ w}}{\text{.r}}{\text{.t}}{\text{. '}}x{\text{';}} \cr & \frac{{df}}{{dx}} = \frac{d}{{dx}}\left( {\frac{1}{{9\ln 7}}} \right)\frac{{\ln x}}{{\sqrt x }} \cr & \,\,\,\,\,\,\,\,\, = \left( {\frac{1}{{9\ln 7}}} \right)\frac{d}{{dx}}\frac{{\ln x}}{{\sqrt x }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\frac{d}{{dx}}cw\left( x \right) = c\frac{d}{{dx}}w\left( x \right)} \right) \cr & \,\,\,\,\,\,\,\,\, = \left( {\frac{1}{{9\ln 7}}} \right)\frac{{\sqrt x \frac{d}{{dx}}\left( {\ln x} \right) - \ln x\frac{d}{{dx}}\sqrt x }}{x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{From quotient rule}}} \right) \cr & \,\,\,\,\,\,\,\,\, = \left( {\frac{1}{{9\ln 7}}} \right)\frac{{\sqrt x \left( {\frac{1}{x}} \right) - \ln x\frac{1}{{2\sqrt x }}}}{x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\frac{d}{{dx}}\left( {\ln x} \right) = \frac{1}{x},\frac{d}{{dx}}{x^n} = n{x^{n - 1}}} \right) \cr & \,\,\,\,\,\,\,\,\, = \left( {\frac{1}{{9\ln 7}}} \right)\frac{{\frac{1}{{\sqrt x }} - \frac{{\ln x}}{{2\sqrt x }}}}{x} \cr & \,\,\,\,\,\,\,\,\, = \left( {\frac{1}{{9\ln 7}}} \right)\frac{{2 - \ln x}}{{2x\sqrt x }} \cr & \cr & \frac{{df}}{{dx}} = \left( {\frac{1}{{9\ln 7}}} \right)\frac{{2 - \ln x}}{{2x\sqrt x }} \cr} {/eq}