Воспользуемся правилом дифференцирования частного:

Получаем:

\(\displaystyle\left(\frac{\color{green}{5x^4-5x+3}}{\color{blue}{\ln (x)}}\right)^{\prime}=\frac{\left(\color{green}{5x^4-5x+3}\right)^{\prime}\left(\color{blue}{\ln (x)}\right)-\left(\color{green}{5x^4-5x+3}\right)\left(\color{blue}{\ln (x)}\right)^{\prime}}{\left(\color{blue}{\ln (x)}\right)^2}{\small .}\)

Вычислим производные:

• \(\displaystyle \begin{aligned} \color{green}{\left(5x^4-5x+3\right)^{\prime}}= \color{green}{\left(5x^4\right)^{\prime}-(5x)^{\prime}+(3)^{\prime}}= \color{green}{\left(20x^3\right)-(5)+(0)} =\color{green}{20x^3-5} {\small,} \end{aligned}\)
• \(\displaystyle \color{blue}{\left(\ln (x)\right)^{\prime}}= \color{blue}{\frac{1}{x}} {\small.} \)

Подставляя, получаем:

\(\displaystyle \begin{aligned}\frac{\color{green}{\left(5x^4-5x+3\right)^{\prime}}\left({\ln (x)}\right)-\left({5x^4-5x+3}\right)\color{blue}{\left(\ln (x)\right)^{\prime}}}{\left({\ln (x)}\right)^2}=\\[10px]=\frac{\color{green}{\left(20x^3-5\right)}\cdot{\ln (x)}-\left({5x^4-5x+3}\right)\cdot\color{blue}{\frac{1}{x}}}{\left({\ln (x)}\right)^2}{\small.}\end{aligned}\)

Таким образом, получаем:

\(\displaystyle \begin{aligned}\left(\frac{{5x^4-5x+3}}{{\ln (x)}}\right)^{\prime}=\frac{\left({5x^4-5x+3}\right)^{\prime}\left({\ln (x)}\right)-\left({5x^4-5x+3}\right)\left({\ln (x)}\right)^{\prime}}{\left({\ln (x)}\right)^2}=\\[10px]=\frac{{\left(20x^3-5\right)}\cdot{\ln (x)}-\left({5x^4-5x+3}\right){\cdot \frac{1}{x}}}{\left({\ln (x)}\right)^2}{\small.}\end{aligned}\)

Ответ: \(\displaystyle \frac{{\left(20x^3-5\right)}\cdot{\ln (x)}-\left({5x^4-5x+3}\right){\cdot \frac{1}{x}}}{\left({\ln (x)}\right)^2}{\small.}\)