Partial Differentiation of Composite Functions

Theorem 4..4 $1. $

Suppose that $z = f(x,y)$

is totally differentiable and $x = x(t), y = y(t)$

are differentiable. Then a compostite function $z = f((x(t),y(t))$

is differentiable and the following is true.

$\displaystyle \frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}.$

Suppose that $z = f(x,y)$

is totally differentiable and $x = x(r,s), y = y(r,s)$

are differentiable. Then a composite function $z = f(x(r,s),y(r,s))$

is also differentiable and the following is true.

$\displaystyle \frac{\partial z}{\partial r} = \frac{\partial z}{\partial x}\fra... ...al x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s}$

NOTE 1. Let $t$ be a time, $x$ be a number of planktons, $y$ be a water temparature, $z$ be a number of minnows. Then $\frac{dz}{dt}$ is an instanteneous change of a number of minnows with respect to a time. But a number of minnows and a number of planktons are influenced by the temparature of the water. $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ . Thus the instanteneous chage of the number of planktons with respect to the temparature is $\frac{dx}{dt}$ and instance change of the water temparature with respect to the time is $\frac{dy}{dt}$ . Thus, the instanteneous change of the number of minnows with respect to the time is

$\displaystyle \frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$

Theorem4.4 2. Express by the tree diagram. 4.2

**Figure 4.2:** Composite Functions
$\includegraphics[width=4.5cm]{SOFTFIG-4/composition.eps}$

Proof 2.

$\displaystyle \frac{\partial z}{\partial r}$	$\displaystyle =$	$\displaystyle \lim_{\Delta r \rightarrow 0} \frac{\Delta z}{\Delta r} = \lim_{\... ...ilon_{1}\frac{\Delta x}{\Delta r} + \varepsilon_{2}\frac{\Delta y}{\Delta r} \}$
	$\displaystyle =$	$\displaystyle \frac{\partial z}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial r} \ensuremath{ \blacksquare}$

Example 4..12 Find $\displaystyle{\frac{dz}{dt}}$
1. $\displaystyle{z = e^{xy^2}, x = t\cos{t}, y = t\sin{t}}$ 2. $z = f(t^2, e^t)$

SOLUTION 1.

$\displaystyle \frac{dz}{dt}$	$\displaystyle =$	$\displaystyle \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$
	$\displaystyle =$	$\displaystyle y^{2}e^{xy^2}(\cos{t} - t\sin{t}) + 2xye^{xy^2}(\sin{t} + t\cos{t}) \ensuremath{ \blacksquare}$

2. Let $x = t^2, y = e^t$ . Then $z = f(x,y)$ and

$\displaystyle \frac{dz}{dt}$

$\displaystyle =$

$\displaystyle \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\p... ...l z}{\partial x} + e^t \frac{\partial z}{\partial y}\ensuremath{ \blacksquare}$

Exercise 4..12 1. Find $\frac{\partial z}{\partial r}, \frac{\partial z}{\partial s}$ of the following function.

$\displaystyle z = \tan^{-1}{\frac{y}{x}}, x = r^3 - 3rs^2, y = 3r^2 s - s^3$

2. For $\displaystyle{z = f(x^{2}y)}$ , Show $\displaystyle{x(\frac{\partial z}{\partial x}) = 2y(\frac{\partial z}{\partial y})}$ .

SOLUTION 1.

$\displaystyle \frac{\partial z}{\partial r}$	$\displaystyle =$	$\displaystyle \frac{\partial z}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial r}$
	$\displaystyle =$	$\displaystyle \frac{1}{1 + (\frac{x}{y})^2}(-\frac{y}{x^2})(3r^2 - 3s^2) + \frac{1}{1 + (\frac{y}{x})^2}(\frac{1}{x})(6rs)$
	$\displaystyle =$	$\displaystyle \frac{-y}{x^2 + y^2}(3r^2 - 3s^2) + \frac{x}{x^2 + y^2}(6rs)\ensuremath{ \blacksquare}$

$\displaystyle \frac{\partial z}{\partial s}$	$\displaystyle =$	$\displaystyle \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \fra... ...\partial y}{\partial s} = \frac{1}{1 + (\frac{y}{x})^{2}}(-\frac{y}{x^2})(-6rs)$
	$\displaystyle +$	$\displaystyle \frac{1}{1 + (\frac{y}{x})^2}(\frac{1}{x})(3r^{2} - 3s^2) = \frac{-y}{x^2 + y^2}(6rs) + \frac{x}{x^2 + y^2}(3r^2 - 3s^2)\ensuremath{ \blacksquare}$

2. Let $x^{2}y = u$ . Then $z = f(u)$ and by the tree diagram,

$\displaystyle \frac{\partial z}{\partial x} = \frac{dz}{du}\frac{\partial u}{\p... ...z}{\partial y} = \frac{dz}{du}\frac{\partial u}{\partial y} = f^{\prime}(u)x^2$

Thus

$\displaystyle x\frac{\partial z}{\partial x} = f^{\prime}(u)2x^{2}y = 2y\frac{\partial z}{\partial y} \ensuremath{ \blacksquare}$

Exercise A

1.

Find $\displaystyle{\frac{dz}{dt}}$ .

(a) $\displaystyle{z = x^{2} + 2y, x = 2t, y = t^{3}}$ (b) $\displaystyle{z = x^{2} + y^{2}, x = \cos{t}, y = \sin{t}}$

(c) $\displaystyle{z = x^{2} + xy + 2y^{2}, x = \cos{t}, y = \sin{t}}$ (d) $\displaystyle{z = x^{3} y^{2}, x = t^{2}, y = t^{3}}$

2.

Find $\displaystyle{\frac{\partial z}{\partial u}, \frac{\partial z}{\partial v}}$ ．

(a) $\displaystyle{z = x^{2} + y^{2}, x = u - 2v, y = 2u + v}$

(b) $\displaystyle{z = x^{2} + xy + 2y^{2}, x = u+v, y = uv}$ (c) $\displaystyle{f(x,y) = x^{2}y^{2}, x = uv, y = v^{2}}$

Exercise B

1.

Find $\displaystyle{\frac{dz}{dt}}$ provided $f$

is in

(a) $\displaystyle{z = \log{(x^2 + y^2)}, x = t + \frac{1}{t}, y = t(t-1)}$ (b) $\displaystyle{z = f(t^2,e^t)}$

2.

Find $\displaystyle{\frac{\partial z}{\partial r}, \frac{\partial z}{\partial s}}$ ．

(a) $\displaystyle{z = \tan^{-1}{\frac{y}{x}}, x = r^3 - 3rs^2, y = 3r^2 s - s^3}$

(b) $\displaystyle{z = \log{\frac{y}{x}}, x = (r-1)^2 + s^2, y = (r+1)^2 + s^2}$

3.

For $\displaystyle{z = f(x,y), x = r\cos{\theta}, y = r\sin{\theta}}$ , show that

$\displaystyle z_{r} = z_{x}\cos{\theta} + z_{y}\sin{\theta}, z_{\theta} = r(-z_{x}\sin{\theta} + z_{y}\cos{\theta}).$