Joseph Krupa

2022-01-15

Derivatives Evaluate the following derivatives.
$\frac{d}{dt}\left({t}^{\frac{1}{t}}\right)$

Thomas White

Expert

Step 1
To evaluate
$\frac{d}{dt}\left({t}^{\frac{1}{t}}\right)$
Step 2
Let $y={t}^{\frac{1}{t}}$
Taking logarithm on both sides
$\mathrm{ln}\left(y\right)=\mathrm{ln}\left({t}^{\frac{1}{t}}\right)$
$\mathrm{ln}\left(y\right)=\frac{1}{t}\mathrm{ln}\left(t\right)$
Differentiate both sides with respect to t
$\frac{1}{y}\frac{dy}{dt}={\left(\frac{1}{t}\right)}^{\prime }\mathrm{ln}\left(t\right)+\left(\frac{1}{t}\right){\left[\mathrm{ln}\left(t\right)\right]}^{\prime }$
$\frac{dy}{dt}=y\left[-\frac{1}{{t}^{2}}\mathrm{ln}\left(t\right)+\frac{1}{t}×\frac{1}{t}\right]$
$\frac{dy}{dt}=y\left[-\frac{1}{{t}^{2}}\mathrm{ln}\left(t\right)+\frac{1}{{t}^{2}}\right]$
$\frac{d}{dt}\left({t}^{\frac{1}{t}}\right)=\left({t}^{\frac{1}{t}}\right)\left[\frac{1}{{t}^{2}}\left(1-\mathrm{ln}\left(t\right)\right)\right]$

Lakisha Archer

Expert

${\left({t}^{\frac{1}{t}}\right)}^{\prime }={\left[{\left({e}^{\mathrm{ln}t}\right)}^{\frac{1}{t}}\right]}^{\prime }={\left({e}^{\left(\frac{1}{t}\right)\cdot \mathrm{ln}t}\right)}^{\prime }=$
$={e}^{\left(\frac{1}{t}\right)\cdot \mathrm{ln}t}\cdot {\left(\frac{1}{t}\mathrm{ln}t\right)}^{\prime }=$
$={e}^{\left(\frac{1}{t}\right)\cdot \mathrm{ln}t}\cdot \left(-\frac{1}{{t}^{2}}\cdot \mathrm{ln}t+\frac{1}{t}\cdot \frac{1}{t}\right)=$
$={t}^{\frac{1}{t}}\cdot \frac{1}{{t}^{2}}\left(-\mathrm{ln}t+1\right)$
$={t}^{\left(\frac{1}{t}\right)-2}\left(1-\mathrm{ln}t\right)$
Result:
${t}^{\left(\frac{1}{t}\right)-2}\left(1-\mathrm{ln}t\right)$

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