Answer to Question #223720 in Calculus for Sarita bartwal

Question

1. f(x,y)= x sin(1/x)+ y sin(1/y)
2. f(x,y,z)= 1/(√ (4-x^2-y^2-z^2)

Answer 1

1 Given that $f(x,y)= x sin(\frac{1}{x})+ y sin(\frac{1}{y})$

$\frac{\partial f(x,y)}{\partial x} = sin(\frac{1}{x}) + x(-\frac{1}{x^2})cos(\frac{1}{x}) = sin(\frac{1}{x}) -\frac{1}{x}cos(\frac{1}{x})$

$\frac{\partial f(x,y)}{\partial y} = sin(\frac{1}{y}) + y(-\frac{1}{y^2})cos(\frac{1}{y}) = sin(\frac{1}{y}) -\frac{1}{y}cos(\frac{1}{y})$

2 Given that $f(x,y,z)= \frac{1}{\sqrt{4-x^2-y^2-z^2}}$

$\frac{\partial f(x,y,z)}{\partial x} = -\frac{-2x}{2({4-x^2-y^2-z^2})^{3/2}} = \frac{x}{({4-x^2-y^2-z^2})^{3/2}}$

$\frac{\partial f(x,y,z)}{\partial y} = -\frac{-2y}{2({4-x^2-y^2-z^2})^{3/2}} = \frac{y}{({4-x^2-y^2-z^2})^{3/2}}$

$\frac{\partial f(x,y,z)}{\partial z} = -\frac{-2z}{2({4-x^2-y^2-z^2})^{3/2}} = \frac{z}{({4-x^2-y^2-z^2})^{3/2}}$