Answer to Question #136839 in Calculus for Promise Omiponle

Question

For the surface (1/x)+(1/y)+(1/z)= 1 evaluate (partial z)/(partial x) and (partial z)/(partial y) at the point (2,3,6)

Answer 1

$$(1/x)+(1/y)+(1/z)= 1$$

Now take partial derivative with respect to x in both sides and using chain rule we get

$$-\frac{1}{x^2}-\frac{1}{z^2}\frac{\partial z}{\partial x}=0\\ \implies \frac{\partial z}{\partial x}=-(\frac{z}{x})^2\\ \implies \frac{\partial z}{\partial x}\bigg|_{(2,3,6)}=-(\frac{6}{2})^2=-9$$

Now, since $(1/x)+(1/y)+(1/z)= 1$ is symmetric in x,y,z thus

$$\frac{\partial z}{\partial y}=-(\frac{z}{y})^2\\ \implies \frac{\partial z}{\partial y}\bigg|_{(2,3,6)}=-(\frac{6}{3})^2=-4$$