Answer
$\frac{df(4p^2)}{dp}=24pe^{4p^2}$
Work Step by Step
$f(t)=3e^t$; $t(p)=4p^2$
$\Longrightarrow$
$f(4p^2)=3e^{4p^2}$
Taking $\ln$
$\ln f(4p^2)=\ln3e^{4p^2}$
$\ln f(4p^2)=\ln3 +\ln e^{4p^2}$
$\ln f(4p^2)=\ln3 +4p^2\ln e$
$\ln f(4p^2)=\ln3 +4p^2$
Taking derivative with respect to p
$\frac{d\ln f(4p^2) }{dp}= \frac{ d( \ln3 +4p^2) }{dp}$
$\frac{1}{ f(4p^2)} \frac{df(4p^2)}{dp}= \frac{d\ln3}{dp}+\frac{d(4p^2)}{dp}$
$\frac{1}{ f(4p^2)} \frac{df(4p^2)}{dp}= 8p$
$ \frac{df(4p^2)}{dp}= { f(4p^2)}8p$
$ \frac{df(4p^2)}{dp}= (3e^{4p^2}) 8p=24pe^{4p^2}$
