Find the derivative g(t) = t^2/1+logt

Expert Answers

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You need to use product rule to find derivative of the given function such that:

`(u/v)' = (u'v - uv')/v^2`

Reasoning by analogy yields:

`g'(t) = ((t^2)'(1+ln t) - (t^2)(1+ln t)')/((1 + ln t)^2)`

`g'(t) = (2t(1+ln t) - (t^2)(1/t))/((1 + ln t)^2)`

Reducing by t yields:

`g'(t) = (2t(1+ln t) - t)/((1 + ln t)^2)`

You need to factor out t such that:

`g'(t) = (t(2(1+ln t) - 1))/((1 + ln t)^2)`

Opening the brackets yields:

`g'(t) = (t(2 + 2ln t - 1))/((1 + ln t)^2)`

`g'(t) = (t(1+ ln t^2))/((1 + ln t)^2) `

Hence, evaluating the derivative of the given function yields `g'(t) = (t(1+ ln t^2))/((1 + ln t)^2).`

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