If f(t) = frac{t}{2} + frac{1}{4} log(2t - 1) | vedclass

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(B) Given $f(t) = \frac{t}{2} + \frac{1}{4} \log(2t - 1)$.First,find the derivative $f^{\prime}(t)$:$f^{\prime}(t) = \frac{d}{dt} \left( \frac{t}{2} +

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$1+t$

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(B) Given $f(t) = \frac{t}{2} + \frac{1}{4} \log(2t - 1)$.First,find the derivative $f^{\prime}(t)$:$f^{\prime}(t) = \frac{d}{dt} \left( \frac{t}{2} + \frac{1}{4} \log(2t - 1) \right) = \frac{1}{2} + \frac{1}{4} \cdot \frac{1}{2t - 1} \cdot 2 = \frac{1}{2} + \frac{1}{2(2t - 1)}$.Now,substitute $t$ with $\frac{t+1}{2t+1}$ in $f^{\prime}(t)$:$f^{\prime}\left(\frac{t+1}{2t+1}\right) = \frac{1}{2} + \frac{1}{2\left(2\left(\frac{t+1}{2t+1}\right) - 1\right)}$.Simplify the denominator term:$2\left(\frac{t+1}{2t+1}\right) - 1 = \frac{2t + 2 - (2t + 1)}{2t + 1} = \frac{1}{2t + 1}$.Substitute this back into the expression:$f^{\prime}\left(\frac{t+1}{2t+1}\right) = \frac{1}{2} + \frac{1}{2\left(\frac{1}{2t+1}\right)} = \frac{1}{2} + \frac{2t+1}{2} = \frac{1 + 2t + 1}{2} = \frac{2t + 2}{2} = t + 1$.

If $f(t) = \frac{t}{2} + \frac{1}{4} \log(2t - 1)$,then $f^{\prime}\left(\frac{t+1}{2t+1}\right) = $

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