note1 - Microeconomic Theory Econ 101A Fall 2008 GSI Eva...

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Microeconomic Theory Econ 101A Fall 2008 GSI: Eva Vivalt Section Notes 1: Calculus and Optimization 1 Multi-Variable Calculus 1.1 Partial Differentiation The partial derivative of a multi-variable function f(x 1 , x 2 ) is the incremental change in the function caused by an incremental change in one of the variables while all other variables are held constant. The formal definition of the partial derivative is: ∂f ( x 1 , x 2 ) ∂x 1 = lim h →∞ f ( x 1 + h, x 2 ) - f ( x 1 , x 2 ) h For convenience we sometimes write partial derivatives using the following alternative notations: ∂f ( x 1 , x 2 ) ∂x 1 = f 1 ( x 1 , x 2 ) = f x 1 ( x 1 , x 2 ) Example 1. Say our function of x and y is 2x 3 y. ∂x 2 x 3 y = 2 y ∂x x 3 = 6 x 2 y ∂y 2 x 3 y = 2 x 3 ∂y y = 2 x 3 1.2 Total Differentiation For small changes in the function f(x 1 , x 2 ), a first order Taylor approximation holds exactly, thus the total change in f is just the sum of the partials for each variable (slopes in each direction) times the incremental change in that particular variable. df ( x 1 , x 2 ) = ∂f ( x 1 , x 2 ) ∂x 1 dx 1 + ∂f ( x 1 , x 2 ) ∂x 2 dx 2 = f 1 ( x 1 , x 2 ) dx 1 + f 2 ( x 1 , x 2 ) dx 2 1
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Now let’s assume that y is a known function of x, y=h(x). We could find the total derivative of f(x,y) with respect to x using the formula above so that: df ( x, y ) dx = df ( x, h ( x )) dx = f 1 ( x, h ( x )) dx dx + f 2 ( x, h ( x )) dh ( x ) dx = f 1 ( x, h ( x )) + f 2 ( x, h ( x )) dh ( x ) dx Example 2. Again, take our function of x and y . By our previous work, df ( x, y ) = ∂f ( x, y ) ∂x dx + ∂f ( x, y ) ∂y dy = 6 x 2 ydx + 2 x 3 dy df ( x, y ) dx = ∂f ( x, y ) ∂x + ∂f ( x, y ) ∂y dy dx And if we know the relationship between x and y, e.g. f ( x ) = y = x 2 then we can note: df ( x ) dx = 2 x And consequently: df ( x, y ) dx = 6 x 2 y + 4 x 4 1.3 Implicit Differentiation Sometimes we may be in a situation where y depends on x but there is no ’explicit’ formula that can be solved to express y in terms of x, yet it is known that x and y satisfy some equation such as: f ( x, y ) = k where k is a constant. Even though we can’t explicitly solve for a function h s.t. y = h(x), it may still be possible to find the derivative dh ( x ) dx 2
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To see this, start by taking the total derivative of the equation above with respect to x. The derivative of the LHS is: df ( x, y ) dx = f 1 ( x, h ( x )) + f 2 ( x, h ( x )) dh ( x ) dx The derivative of the RHS is: dk dx = 0 thus we can rearrange and solve: f 1 ( x, h ( x )) + f 2 ( x, h ( x )) dh ( x ) dx = 0 dh ( x ) dx = - f 1 ( x, h ( x )) f 2 ( x, h ( x )) which is defined as long as f 2 ( x, h ( x )) 6 = 0 .
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