2Section Outline Partial Derivatives Computing Partial Derivatives Evaluating Partial Derivatives at a PointLocal Approximation of f (x, y)Demand EquationsSecond Partial Derivative
3Partial Derivatives Definition Example Partial Derivative of f (x, y) with respect to x: Written ,the derivative of f (x, y), where y is treated as a constant and f (x, y) is considered as a function of x aloneIf , then
4Computing Partial Derivatives EXAMPLECompute forSOLUTIONTo compute , we only differentiate factors (or terms) that contain x andwe interpret y to be a constant.This is the given function.Use the product rule where f (x) = x2 and g(x) = e3x.To compute , we only differentiate factors (or terms) that contain y andwe interpret x to be a constant.
5Computing Partial Derivatives CONTINUEDThis is the given function.Differentiate ln y.
6Computing Partial Derivatives EXAMPLECompute forSOLUTIONTo compute , we treat every variable other than L as a constant. ThereforeThis is the given function.Rewrite as an exponent.Bring exponent inside parentheses.Note that K is a constant.Differentiate.
7Evaluating Partial Derivatives at a Point EXAMPLELet Evaluate at (x, y, z) = (2, -1, 3).SOLUTION
9Local Approximation of f (x, y) EXAMPLELet Interpret the resultSOLUTIONWe showed in the last example thatThis means that if x and z are kept constant and y is allowed to vary near -1, then f (x, y, z) changes at a rate 12 times the change in y (but in a negative direction). That is, if y increases by one small unit, then f (x, y, z) decreases by approximately 12 units. If y increases by h units (where h is small), then f (x, y, z) decreases by approximately 12h. That is,
10Demand EquationsEXAMPLEThe demand for a certain gas-guzzling car is given by f (p1, p2), where p1 isthe price of the car and p2 is the price of gasoline. Explain whySOLUTIONis the rate at which demand for the car changes as the price of the carchanges. This partial derivative is always less than zero since, as the price of the car increases, the demand for the car will decrease (and visa versa).is the rate at which demand for the car changes as the price of gasolinechanges. This partial derivative is always less than zero since, as the price of gasoline increases, the demand for the car will decrease (and visa versa).
11Second Partial Derivative EXAMPLELet FindSOLUTIONWe first note that This means that to compute , wemust take the partial derivative of with respect to x.