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Differentiate. f(t)= square root of (lnt+t)
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@Leader can you help? lol sorry for keep tagging you
Sorry don't know this...
Is it \(\sqrt{\ln(t) + t}\)?
Yea, differentiate it ^
Do these: \(f(x) = \sqrt{x}\) \(g(x) = \ln(x)\) \(h(x) = x\) Find \(f'(x),\;g'(x),\;and\;h'(x)\)
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\[\frac{ t+1 }{ 2t \sqrt{\ln t +t}}\]
\[f(t)=\sqrt{\ln(t)+t}\] \[f(t)=(\ln(t)+t)^{1/2}\] \[f'(t)=\frac{ 1 }{ 2 }(\ln(t)+t)^{-1/2}\times(\frac{ 1 }{ t }+1)\] \[f'(t)=\frac{ 1 }{ 2\sqrt{\ln(t)+t} }\times(\frac{ 1 }{ t }+1)\] \[f'(t)=\frac{ 1+t }{ 2t \sqrt{\ln(t)+t} }\]
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