Answered: "Derivatives Evaluate the following derivatives. ddt(t1t)"

Joseph Krupa

Answered question

2022-01-15

Derivatives Evaluate the following derivatives.

\frac{d}{d t} (t^{\frac{1}{t}})

Thomas White

Beginner2022-01-16Added 40 answers

Step 1
To evaluate

\frac{d}{d t} (t^{\frac{1}{t}})

Step 2
Let

y = t^{\frac{1}{t}}

Taking logarithm on both sides

\ln (y) = \ln (t^{\frac{1}{t}})

\ln (y) = \frac{1}{t} \ln (t)

Differentiate both sides with respect to t

\frac{1}{y} \frac{d y}{d t} = {(\frac{1}{t})}^{'} \ln (t) + (\frac{1}{t}) {[\ln (t)]}^{'}

\frac{d y}{d t} = y [- \frac{1}{t^{2}} \ln (t) + \frac{1}{t} \times \frac{1}{t}]

\frac{d y}{d t} = y [- \frac{1}{t^{2}} \ln (t) + \frac{1}{t^{2}}]

\frac{d}{d t} (t^{\frac{1}{t}}) = (t^{\frac{1}{t}}) [\frac{1}{t^{2}} (1 - \ln (t))]

Lakisha Archer

Beginner2022-01-17Added 39 answers

{(t^{\frac{1}{t}})}^{'} = {[{(e^{\ln t})}^{\frac{1}{t}}]}^{'} = {(e^{(\frac{1}{t}) \cdot \ln t})}^{'} =

= e^{(\frac{1}{t}) \cdot \ln t} \cdot {(\frac{1}{t} \ln t)}^{'} =

= e^{(\frac{1}{t}) \cdot \ln t} \cdot (- \frac{1}{t^{2}} \cdot \ln t + \frac{1}{t} \cdot \frac{1}{t}) =

= t^{\frac{1}{t}} \cdot \frac{1}{t^{2}} (- \ln t + 1)

= t^{(\frac{1}{t}) - 2} (1 - \ln t)

Result:

t^{(\frac{1}{t}) - 2} (1 - \ln t)

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