Chapter 15 - Differentiation in Several Variables - 15.4 Differentiability and Tangent Planes - Exercises - Page 789: 2

$$z= -0.3378+0.966(y-0.8)$$

Given $$f(x, y)=0.2 x^{4}+ y^{6}-x y$$ Since \begin{align*} f(x,y)&= 0.2 x^{4}+y^{6}-x y\ \ \ \ & f(1,0.8)&=-0.3378\\ f_x(x,y)&= 0.8 x^{3} -y \ \ \ \ & f_x(1,0.8)&= 0\\ f_y(x,y)&= 6y^5-x \ \ \ \ & f_y(1,0.8)&= 0.966\\ \end{align*} Then the tangent plane at $(1,0.8)$ is given by \begin{align*} z&= f(1,0.8)+f_{x}(1,0.8)(x-1)+f_{y}(1,0.8)(y-0.8)\\ &= -0.3378+0.966(y-0.8) \end{align*}