Chapter 15 - Differentiation in Several Variables - 15.4 Differentiability and Tangent Planes - Exercises - Page 789: 5

$$z=17+8(x-4)-2(y-1)$$

Given $$f(x, y)= x^2+y^{-2} \quad \text { at } \quad(4,1)$$ Since \begin{aligned} f_{x}(x, y) &=2x \\ f_{x}(4,1) &= 8 \end{aligned} and \begin{aligned} f_{y}(x, y) &=-2y^{-3}\\ f_{y}(4,1) &=-2 \end{aligned} Then the tangent plane given by \begin{aligned} z &=f(4,1)+f_{x}(4,1)(x-4)+f_{y}(4,1)(y-1) \\ &=17+8(x-4)-2(y-1) \end{aligned}