What happened to all of the responses you had just a few minutes ago?
If E is a function of p and r and p and r themselves are functions of t the "chain rule" says that
\(\displaystyle \frac{dE}{dt}= \frac{\partial E}{\partial p}\frac{dp}{dt}+ \frac{\partial E}{\partial r}\frac{dr}{dt}\).
In this problem \(\displaystyle E(p, r)= Ap^{-a}r^b\) so \(\displaystyle \frac{\partial E}{\partial p}= -aAp^{-a-1}r^b\) and \(\displaystyle \frac{\partial E}{\partial r}= bAp^{-a}t=r^{b-1}\).
Further \(\displaystyle p(t)= p_0(1.06)^t\). Since this problem is about "logarithmic differentiation" we can write this as \(\displaystyle ln(p_0)+ tln(1.06)\) and then, differentiating both sides, \(\displaystyle \frac{1}{p}\frac{dp}{dt}= ln(1.06)\) so that \(\displaystyle \frac{dp}{dt}= ln(1.06)p= p_0ln(1.06)(1.06)^t\).
Similarly \(\displaystyle r(t)= r_0(1.08)^t\) so that \(\displaystyle ln(r)= ln(r_0)+ t ln(1.08)\), \(\displaystyle \frac{1}{r}\frac{dr}{dt}= ln(1.08)\), and \(\displaystyle \frac{dr}{dt}= ln(1.08)r= r_0ln(1.08)(1.08)^t\).
I'll leave it to you to put those together in
If E is a function of p and r and p and r them selves are functions of t the "chain rule" says that
\(\displaystyle \frac{dE}{dt}= \frac{\partial E}{\partial p}\frac{dp}{dt}+ \frac{\partial E}{\partial r}\frac{dr}{dt}\).
\(\displaystyle \frac{dE}{dt}= \frac{\partial E}{\partial p}\frac{dp}{dt}+ \frac{\partial E}{\partial r}\frac{dr}{dt}\).
In this problem \(\displaystyle E(p, r)= Ap^{-a}r^b\) so \(\displaystyle \frac{\partial E}{\partial p}= -aAp^{-a-1}r^b\) an \(\displaystyle \frac{\partial E}{\partial r}= bAp^{-a}t=r^{b-1}\).
Further \(\displaystyle p(t)= p_0(1.06)^t\). Since this problem is about "logarithmic differentiation" we can write this as \(\displaystyle ln(p_0)+ tln(1.06)\) and then, differentiating both sides, \(\displaystyle \frac{1}{p}\frac{dp}{dt}= ln(1.06)\) so that \(\displaystyle \frac{dp}{dt}= ln(1.06)p= p_0ln(1.06)(1.06)^t\).
Similarly \(\displaystyle r(t)= r_0(1.08)^t\) so that \(\displaystyle ln(r)= ln(r_0)+ t ln(1.08)\), \(\displaystyle \frac{1}{r}\frac{dr}{dt}= ln(1.08)\), and \(\displaystyle \frac{dr}{dt}= ln(1.08)r= r_0ln(1.08)(1.08)^t\).
