Blog: `h(t) = ln(t)/t` Find the derivative of the function.

$\displaystyle{h}{\left({t}\right)}=\frac{{\ln{{\left({t}\right)}}}}{{t}}$ Find the derivative of the function.

We'll use the formula of the derivative of a quotient: $\displaystyle{\left(\frac{{u}}{{v}}\right)}'=\frac{{{u}'{v}-{u}{v}'}}{{{v}}^{{2}}}.$ Here u= $\displaystyle{\ln{{\left({t}\right)}}}$ and $\displaystyle{v}={t},$ therefore $\displaystyle{u}'=\frac{{1}}{{t}}$ and $\displaystyle{v}'={1}.$ So

$\displaystyle{h}'{\left({t}\right)}=\frac{{\frac{{1}}{{t}}\cdot{t}-{\ln{{\left({t}\right)}}}}}{{{t}}^{{2}}}=\frac{{{1}-{\ln{{\left({t}\right)}}}}}{{{t}}^{{2}}}.$

We'll use the formula of the derivative of a quotient: $\displaystyle{\left(\frac{{u}}{{v}}\right)}'=\frac{{{u}'{v}-{u}{v}'}}{{{v}}^{{2}}}.$ Here u= $\displaystyle{\ln{{\left({t}\right)}}}$ and $\displaystyle{v}={t},$ therefore $\displaystyle{u}'=\frac{{1}}{{t}}$ and $\displaystyle{v}'={1}.$ So

$\displaystyle{h}'{\left({t}\right)}=\frac{{\frac{{1}}{{t}}\cdot{t}-{\ln{{\left({t}\right)}}}}}{{{t}}^{{2}}}=\frac{{{1}-{\ln{{\left({t}\right)}}}}}{{{t}}^{{2}}}.$

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$\displaystyle{h}{\left({t}\right)}=\frac{{\ln{{\left({t}\right)}}}}{{t}}$ Find the derivative of the function.

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$\displaystyle{h}{\left({t}\right)}=\frac{{\ln{{\left({t}\right)}}}}{{t}}$ Find the derivative of the function.

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