How to solve an exercise with supply and demand (derivatives)? Supply and Demand. Supply of a firm is given by S = $e^p$ . Demand is given by $A/P$ .Find the equilibrium condition (supply and demand must be equal). How does the price change when the demand function increases to A'?
Hint: You can use implicit differentiation in the equilibrium condition. Alternatively, you can solve for A and differentiate the inverse function. Once you have an expression, try to guess its sign based on what you know about exponential functions.
In the image, you see my solution (hyperlink). But I do not know what to do next. Please, help me)
Edit: Solution in text:
$Q^S$ = $e^P$
$Q^D = A÷P$
$e^P$ = A÷P
A = P$e^P$
A' = (P$e^P$)' = $e^P$ + P$e^P$
$e^P$ ( 1 + P ) = 0
1 + P = 0
P = -1
P=-1 is a minimum point
$Q^D$ = $A/P$ + $e^P$ (1+P)
my solution. click here

Try this. The Equilibrium Condition is $Q^S=Q^D$ or $e^P=A/P$ or $$Pe^P=A.$$  We want $\frac{\partial P}{\partial A}.$ Re-write the equilibrium condition as $Pe^P-A=0$ and use implicit differentiation, if $g(x,k)=0$ then $$\frac{dx}{dk}=-\frac{\frac{\partial g}{\partial k}}{\frac{\partial g}{\partial x}}.$$
If we let $g$ be $Pe^P-A$ we find that
$$\frac{dP}{dA}=\frac{1}{e^P(1+P)}$$ and since prices are non-negative, the derivative is positive, so  higher $A$ means a higher price, which makes economic sense: higher demand increases the market price.
