Derivative of the function using logs
f(t)=e^2t_ln(t+1) and f(x)=ln(2x/x+1)
Hello factory1o1I assume you mean $\displaystyle f(t) = e^{2t}\ln(t+1)$
in which case, use the product rule to get:
$\displaystyle f'(t)=e^{2t}\times\frac{1}{t+1}+ 2e^{2t}\ln(t+1)$
$\displaystyle =e^{2t}\Big(\frac{1}{t+1}+ 2\ln(t+1)\Big)$
To differentiate the second function, use the properties of logs first:
$\displaystyle f(x) = \ln\left(\frac{2x}{x+1}\right)$
$\displaystyle = \ln(2)+\ln(x) -\ln(x+1)$
$\displaystyle \Rightarrow f'(x) =\frac{1}{x}-\frac{1}{x+1}$
$\displaystyle =\frac{1}{x(x+1)}$
Grandad