Answer to Question #105737 in Calculus for Mujtaba

Question

(a) Suppose that z = f(u) and u = g(x, y). Draw a tree diagram, and use it to construct chain
rules that express ∂z/∂x and ∂z/∂y in terms of dz/du ,
∂u/∂x, and ∂u/∂y .
(b) Let z = f(x ^2 − y^2). Use the result in part (a) to show that
y∂z/∂x + x∂z/∂y = 0
(c) Suppose yz = ln(x + z). Use implicit differentiation to find ∂z/∂x and ∂z/∂y

Answer 1

(a) Diagram

Thus, to obtain ∂z/∂x and ∂z/∂y one should differentiate  u = g(x, y).

$\dfrac{\partial z}{\partial x} = \dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial x}\\ \dfrac{\partial z}{\partial y} = \dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial y}\\$

(b)

$\dfrac{\partial z}{\partial x} = \dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial x} = f'\cdot 2x\\ \dfrac{\partial z}{\partial y} = \dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial y} = -f'\cdot 2y\\$

$y\dfrac{\partial z}{\partial x} + x\dfrac{\partial z}{\partial y} = 2xyf' - 2xyf' = 0.$

QED

(c)$yz = \ln(x + z)$ .

Differentiate both sides with respect to x: $y\dfrac{\partial z}{\partial x} = \dfrac{1}{x+z}(\dfrac{\partial z}{\partial x} + 1).\\$

Express derivative:

$\dfrac{\partial z}{\partial x}(y - \dfrac{1}{x+z}) = \dfrac{1}{x+z}\\ \dfrac{\partial z}{\partial x} = \dfrac{1}{y(x+z) - 1}$

Differentiate both sides with respect to y: $y\dfrac{\partial z}{\partial y} +z = \dfrac{1}{x+z}\dfrac{\partial z}{\partial y} .\\$

Express derivative:

$\dfrac{\partial z}{\partial y} = -\dfrac{1}{y - \dfrac{1}{x+z}} = \dfrac{z(x+z)}{1 - y(x+z)}.\\$

Answer.

a)

$\dfrac{\partial z}{\partial x} = \dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial x}\\ \dfrac{\partial z}{\partial y} = \dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial y}\\$

b)

$y\dfrac{\partial z}{\partial x} + x\dfrac{\partial z}{\partial y} = 0$

c)

$\dfrac{\partial z}{\partial x} = \dfrac{1}{y(x+z) - 1}$

$\dfrac{\partial z}{\partial y} = \dfrac{z(x+z)}{1 - y(x+z)}$