# Find dy/dx. y = 9 ln x + 2 log_9 x

## Question:

Find {eq}\frac{\displaystyle dy}{\displaystyle dx} {/eq}.

{eq}y = 9 \ln x + 2 \log_9 x {/eq}

We are given:

{eq}\displaystyle y = 9 \ln x + 2 \log_9 x {/eq}

Apply logarithmic property {eq}\displaystyle \log_a x = \dfrac{ \ln x}{\ln a } {/eq}

{eq}\displaystyle y = 9 \ln x + \dfrac{ 2\ln x}{\ln 9 } {/eq}

{eq}\displaystyle y = \left( 9 + \dfrac{ 2}{\ln 9 } \right) \ln x {/eq}

Differentiating both sides, we'll get:

{eq}\Rightarrow \displaystyle \dfrac{\ dy}{ \ dx}= \dfrac{\ d }{ \ dx} \left( \left( 9 + \dfrac{ 2}{\ln 9 } \right) \ln x \right). {/eq}

Take the constant out:

{eq}\Rightarrow \displaystyle \dfrac{\ dy}{ \ dx}= \left( 9 + \dfrac{ 2}{\ln 9 } \right) \dfrac{\ d }{ \ dx} \left( \ln x\right). {/eq}

{eq}\displaystyle \Rightarrow \dfrac{\ dy}{ \ dx}= \left( 9 + \dfrac{ 2}{\ln 9 } \right) \dfrac{1}{x} {/eq}

{eq}\displaystyle \Rightarrow \dfrac{\ dy}{ \ dx}= \dfrac{9}{x} + \dfrac{ 2}{x \ln 9 } {/eq}

Therefore the solution is: {eq}\displaystyle {\boxed{ \dfrac{\ dy }{ \ dx } = \dfrac{9}{x} + \dfrac{ 2}{x \ln 9 } .}} {/eq}