Suppose that $f$ is differentiable at $x$, that is, we have the following approximation
$$f'(x) = \lim_{\Delta \to 0}\frac{\Delta f}{\Delta x} = \frac{f(x + \Delta x) - f(x)}{\Delta x}.$$Hence we have
Suppose that $f$ is differentiable at $x$. Then we have $$\frac{dy}{dx} = f'(x) = \frac{\Delta f}{\Delta x} + \varepsilon$$ where $\varepsilon = \varepsilon(x) \to 0$ as $\Delta x \to 0$ or $$\frac{\Delta f}{\Delta x} \approx \frac{dy}{dx} = f'(x), \quad \text{or} \quad \Delta f(x) \approx f'(x)\,\Delta x,$$ whenever $\Delta x$ is sufficiently small. We write $dx = \Delta x$. We define $dy$, at $x = a$, by $$dy = f'(a)\,dx = \frac{dy}{dx}\bigg|_{x=a} \cdot dx,$$ where $dy = dy(a)$, $dx = \Delta x$. For a general $x$, we write $$dy = f'(x)\,dx = \frac{dy}{dx} \cdot dx.$$ The $dy$, $dx$ are called the differentials of $f$ at $x$.
A function of two variables is a rule that assigns to each pair $(x, y)$ of real numbers in a domain $D$ as a subset in the $xy$-plane a unique real number denoted by $f(x, y)$. The range of $f$ is given by $$R = \{f(x, y) \mid (x, y) \in D\}.$$ The collection of points $(x, y, z) = (x, y, f(x, y))$ in $\mathbb{R}^3$ for $(x, y)$ in $D$ constitutes 3-dimensional graph (surface) of $f$.
Example. Find the domain and range of
$$g(x, y) = \sqrt{10 - x^2 - y^2}.$$It is easy to see that the domain of $g$ is given by
$$\{(x, y) :\; x^2 + y^2 \leq 10 = (\sqrt{10})^2\}.$$The range is therefore $\{z :\; 0 \leq z \leq \sqrt{10}\}$.
Example. Find the domain and range of
$$f(x, y) = 8 - 2x - y.$$It is easy to see that the domain of $f$ is the whole $xy$-plane, and the range is whole real-axis $\mathbb{R}$.
Example. Find the domain of
$$f(x, y) = \frac{xy}{x^2 + y^2}.$$It is easy to see that the domain of $f$ is the whole $xy$-plane minus the point $(0, 0)$. But the range is not so easy to find.
Example (Stewart). How do we find the rate of changes of
$$f(x, y) = 4 - x^2 - y^2$$That are, we need to compute, assuming the general $y$ initially:
$$\lim_{h \to 0}\frac{f(1, y+h) - f(1, y)}{h} = \lim_{h \to 0}\frac{4 - 1^2 - (y+h)^2 - (4 - 1^2 - y^2)}{h}$$ $$= \lim_{h \to 0}\frac{y^2 - (y+h)^2}{h} = \lim_{h \to 0}\frac{-2hy - h^2}{h} = \lim_{h \to 0} -2y - h = -2y.$$Hence the derivative equals $-2(1) = -2$ when $x = 1$ and $y = 1$ with respect to $y$. Clearly we could “move” the plane $x = 1$ to another plane parallel to the $y$-axis.
Similarly, we have
$$\lim_{h \to 0}\frac{f(x+h, 1) - f(x, 1)}{h} = \lim_{h \to 0}\frac{4 - (x+h)^2 - 1^2 - (4 - x^2 - 1^2)}{h}$$ $$= \lim_{h \to 0}\frac{x^2 - (x+h)^2}{h} = \lim_{h \to 0}\frac{-2hx - h^2}{h} = \lim_{h \to 0} -2x - h = -2x.$$Hence the derivative equals $-2(1) = -2$ when $y = 1$ and $x = 1$ with respect to $x$. Clearly we could “move” the plane $y = 1$ to another plane parallel to the $x$-axis.
More generally, we define
Let $z = f(x, y)$ be defined on $D$. We define $$f_x(x, y) = \lim_{h \to 0}\frac{f(x+h,\, y) - f(x,\, y)}{h}$$ to be the partial derivative of $f$ with respect to $x$, and $$f_y(x, y) = \lim_{h \to 0}\frac{f(x,\, y+h) - f(x,\, y)}{h}$$ to be the partial derivative of $f$ with respect to $y$, provided that the limits exist.
Other notation includes:
Let $z = f(x, y)$ be defined on $D$. We write, $$f_x(x, y) = \frac{\partial f}{\partial x} = \frac{\partial f}{\partial x}(x, y) = D_x f;$$ $$f_y(x, y) = \frac{\partial f}{\partial y} = \frac{\partial f}{\partial y}(x, y) = D_y f$$ provided that the limits exist.
Example (revisited). Consider
$$f(x, y) = 4 - x^2 - y^2.$$Then
$$f_x = f_x(x, y) = -2x, \qquad f_y = f_y(x, y) = -2y.$$At $(1, 2)$,
$$f_x(1, 2) = -2, \qquad f_y(1, 2) = -4.$$We recall that the equations
Suppose that $f$ has continuous partial derivatives. An equation of the tangent plane to the surface $z = f(x, y)$ passing through the point $(x_0, y_0, z_0)$ is given by
Moreover, $$dz = f_x(x, y)\,dx + f_y(x, y)\,dy = \frac{\partial z}{\partial x}dx + \frac{\partial z}{\partial y}dy$$ is called the total differential.
Example (revisited). Find the tangent plane equation for the function
$$f(x, y) = 4 - x^2 - y^2$$at $(1, 2)$.
Since $f_x(1, 2) = -2$, $\;\;f_y(1, 2) = -4$, and $f(1,2) = 4 - 1 - 4 = -1$:
We recall that if $x = g(t)$ and $y = f(x)$, then Newton's chain rule states that
$$\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt}.$$By taking “limit” of the above tangent plane approximation formula, one can obtain
Suppose that $z = f(x, y)$ has partial derivatives with respect to both $x$ and $y$, where $x = x(t)$, $y = y(t)$. Then $$\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}.$$
Example (revisited). Suppose $x = x(t)$ and $y = y(t)$ in
$$f(x, y) = 4 - x^2 - y^2.$$Then
$$\frac{dz}{dt} = \frac{df}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt} = -2x\frac{dx}{dt} - 2y\frac{dy}{dt}.$$More general formulae also follow:
Suppose that $z = f(x, y)$ has partial derivatives with respect to both $x$ and $y$, where $x = g(s, t)$, $y = h(s, t)$. Then $$\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial s}.$$ and $$\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial t}.$$
Given $f(x, y) = x^2 y + 3xy^2$, compute partial derivatives and evaluate at a point.
Find $f_x(x, y)$.
Now find $f_y(x, y)$ for the same function.
Find $f_x(1, 2)$.
Apply the tangent plane formula and generalised chain rule.
Find the tangent plane to $z = x^2 + y^2$ at the point $(1, 1, 2)$.
If $z = x^2 + y^2$, $x = t$, $y = t^2$, what is $dz/dt$?
Evaluate $dz/dt$ at $t = 1$.
— End of Partial Derivatives Notes —