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§ 7.2

§ 7.2. Partial Derivatives. Section Outline. Partial Derivatives Computing Partial Derivatives Evaluating Partial Derivatives at a Point Local Approximation of f ( x , y ) Demand Equations Second Partial Derivative. Partial Derivatives. Computing Partial Derivatives. EXAMPLE.

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§ 7.2

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  1. §7.2 Partial Derivatives

  2. Section Outline • Partial Derivatives • Computing Partial Derivatives • Evaluating Partial Derivatives at a Point • Local Approximation of f(x, y) • Demand Equations • Second Partial Derivative

  3. Partial Derivatives

  4. Computing Partial Derivatives EXAMPLE Compute for SOLUTION To compute , we only differentiate factors (or terms) that contain x and we 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 and we interpret x to be a constant.

  5. Computing Partial Derivatives CONTINUED This is the given function. Differentiate ln y.

  6. Computing Partial Derivatives EXAMPLE Compute for SOLUTION To compute , we treat every variable other than L as a constant. Therefore This is the given function. Rewrite as an exponent. Bring exponent inside parentheses. Note that K is a constant. Differentiate.

  7. Evaluating Partial Derivatives at a Point EXAMPLE Let Evaluate at (x, y, z) = (2, -1, 3). SOLUTION

  8. Local Approximation of f(x, y)

  9. Local Approximation of f(x, y) EXAMPLE Let Interpret the result SOLUTION We showed in the last example that This 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,

  10. Demand Equations EXAMPLE The demand for a certain gas-guzzling car is given by f (p1, p2), where p1 is the price of the car and p2 is the price of gasoline. Explain why SOLUTION is the rate at which demand for the car changes as the price of the car changes. 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 gasoline changes. 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).

  11. Second Partial Derivative EXAMPLE Let . Find SOLUTION We first note that This means that to compute , we must take the partial derivative of with respect to x.

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