Find derivatives for the functions. Assume a, b, c, and k are constants. f(x)=\frac{x^{3}}{9}(3\ln x-1)

Find derivatives for the functions. Assume a, b, c, and k are constants.
$$\displaystyle{f{{\left({x}\right)}}}={\frac{{{x}^{{{3}}}}}{{{9}}}}{\left({3}{\ln{{x}}}-{1}\right)}$$

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Bentley Leach
$$\displaystyle{f{{\left({x}\right)}}}={\frac{{{x}^{{{3}}}}}{{{9}}}}{\left({3}{\ln{{x}}}-{1}\right)}$$
Differentiate using the product rule
$$\displaystyle{f}'{\left({x}\right)}={\left({\frac{{{x}^{{{3}}}}}{{{9}}}}\right)}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{3}{\ln{{x}}}-{1}\right]}+{\left({3}{\ln{{x}}}-{1}\right)}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{\frac{{{x}^{{{3}}}}}{{{9}}}}\right]}$$
Therefore,
$$\displaystyle{f}'{\left({x}\right)}={\left({\frac{{{x}^{{{3}}}}}{{{9}}}}\right)}{\left({\frac{{{3}}}{{{x}}}}\right)}+{\left({3}{\ln{{x}}}-{1}\right)}{\left({\frac{{\lbrace}}{{3}}}{x}^{{{2}}}{\left\lbrace{9}\right\rbrace}\right)}$$
Simplify
$$\displaystyle{f}'{\left({x}\right)}={\frac{{{x}^{{{2}}}}}{{{3}}}}+{x}^{{{2}}}{\ln{{x}}}-{\frac{{{x}^{{{2}}}}}{{{3}}}}$$
$$\displaystyle{f}'{\left({x}\right)}={x}^{{{2}}}{\ln{{x}}}$$