Partial Differentiation of Composite Functions

$\partial$Symbol

$d$ is used for the sign of derivative of a function of single variable. $\partial$ is used for the sign of derivative of a function of more than one variables.

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.

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

$2.$ 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.

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

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

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

Theorem4.4 2. Express by the tree diagram. 4.22

Tree Diagram

$z$ is a function of $x$ and $y$. Then draw a line from $z$ to $x$, and $z$ to $y$. Now $x$ and $y$ are functions of $r$ and $s$. Then draw lines from $x$ and $y$ to $r$ and $s$.

Figure 4.22: Composite Functions

Figure 4.21: Composite function

Proof 2.

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

$$= \frac{\partial z}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial r} \ensuremath{ \blacksquare}$$

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

SOLUTION 1.

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

$$= y^{2}e^{xy^2}(\cos{t} - t\sin{t}) + 2xye^{xy^2}(\sin{t} + t\cos{t}) \ensuremath{ \blacksquare}$$

$z$ is a function of $x$ and $y$, $x$ and $y$ are functions of $t$. Then draw a line from $z$ to $x,y$ and lines from $x$ to $t$, from $y$ to $t$.

Figure 4.23: Example4-12-1

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

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

Since $z = f(x,y)$ is not known, $\frac{\partial z}{\partial x}$ is the final form

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

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

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

Figure 4.24: Exercise4-12-1

SOLUTION 1.

$$\frac{\partial z}{\partial r} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial r}$$

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

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

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

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

Figure 4.25: Exercise4-12-2

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

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

Thus

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

Exercise A

1.

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

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

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

2.

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

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

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

Exercise B

1.

Find $${\frac{dz}{dt}}$$ provided $f$ is in $C^(1)$.

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

(c) $${z = f(2t, 4t^2)}$$ (d) $${z = x^2 - 2y^2, x = \cos{t}, y = \sin{t}}$$

2.

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

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

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

(c) $${z = \sqrt{x^{2} + y^{2}}, x = r\cos{s}, y = r\sin{s}, (r > 0)}$$

3.

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

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