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PartialDifferentiation

# PartialDifferentiation - CHAPTER 7 Partial Dierentiation...

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CHAPTER 7 Partial Differentiation From the previous two chapters we know how to differentiate functions of one variable . But many functions in economics depend on several variables: output depends on both capital and labour, for example. In this chapter we show how to differentiate functions of several vari- ables and apply this to find the marginal products of labour and capital, marginal utilities, and elasticities of demand . We use differentials to find the gradients of isoquants, and hence determine marginal rates of substitution . 1. Partial Derivatives The derivative of a function of one variable, such as y ( x ), tells us the gradient of the function: how y changes when x increases. If we have a function of more than one variable, such as: z ( x, y ) = x 3 + 4 xy + 5 y 2 we can ask, for example, how z changes when x increases but y doesn’t change. The answer to this question is found by thinking of z as a function of x , and differentiating, treating y as if it were a constant parameter: ∂z ∂x = 3 x 2 + 4 y This process is called partial differentiation. We write ∂z ∂x rather than dz dx , to emphasize that z is a function of another variable as well as x , which is being held constant. ∂z ∂x is called the partial derivative of z with respect to x ∂z ∂x is pronounced “partial dee z by dee x”. Similarly, if we hold x constant, we can find the partial derivative with respect to y : ∂z ∂y = 4 x + 10 y Remember from Chapter 4 that you can think of z ( x, y ) as a “surface” in 3 dimensions. Imagine that x and y represent co-ordinates on a map, and z ( x, y ) represents the height of the land at the point ( x, y ). Then, ∂z ∂x tells you the gradient of the land as you walk in the direction of increasing x , keeping the y co-ordinate constant. If ∂z ∂x > 0, you are walking uphill; if it is negative you are going down. (Try drawing a picture to illustrate this.) 117

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118 7. PARTIAL DIFFERENTIATION Exercises 7.1 : Find the partial derivatives with respect to x and y of the functions: (1) f ( x, y ) = 3 x 2 - xy 4 (3) g ( x, y ) = ln x y (2) h ( x, y ) = ( x + 1) 2 ( y + 2) Examples 1.1 : For the function f ( x, y ) = x 3 e - y : (i) Show that f is increasing in x for all values of x and y . ∂f ∂x = 3 x 2 e - y Since 3 x 2 > 0 for all values of x , and e - y > 0 for all values of y , this derivative is always positive: f is increasing in x . (ii) For what values of x and y is the function increasing in y ? ∂f ∂y = - x 3 e - y When x < 0 this derivative is positive, so f is increasing in y . When x > 0, f is decreasing in y . 1.1. Second-order Partial Derivatives For the function in the previous section: z ( x, y ) = x 3 + 4 xy + 5 y 2 we found: ∂z ∂x = 3 x 2 + 4 y ∂z ∂y = 4 x + 10 y These are the first-order partial derivatives. But we can differentiate again to find second- order partial derivatives. The second derivative with respect to x tells us how ∂z ∂x changes as x increases, still keeping y constant.
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PartialDifferentiation - CHAPTER 7 Partial Dierentiation...

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