09 Partial derivatives, part 1

Partial derivatives of functions \(\mathbb{R}^2 \to \mathbb{R}\)

In the following definition, recall the two standard basis vectors of \(\mathbb{R}^2\): \[ \mathbf{e}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \quad \text{and} \quad \mathbf{e}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}. \]
Remember that \(\mathbf{e}_1\) is a unit vector that points along the \(x\)-axis, while \(\mathbf{e}_2\) is a unit vector that points along the \(y\)-axis.

Definition.

Let \(f:\mathbb{R}^2 \to \mathbb{R}\) be a differentiable function and \((x_0,y_0)\) fixed point in \(\mathbb{R}^2\).

The partial derivative of \(f\) with respect to \(x\) at \((x_0,y_0)\) is defined as \[ \frac{\partial f}{\partial x}(x_0,y_0) = f'(x_0,y_0)\mathbf{e}_1. \]
The partial derivative of \(f\) with respect to \(y\) at \((x_0,y_0)\) is defined as \[ \frac{\partial f}{\partial y}(x_0,y_0) = f'(x_0,y_0)\mathbf{e}_2. \]
If we replace \((x_0,y_0)\) with a variable point \((x,y)\), we get the partial derivative functions \[ z = \frac{\partial f}{\partial x}(x,y) \quad \text{and} \quad z = \frac{\partial f}{\partial y}(x,y). \]

What do partial derivatives measure?

The partial derivative \(\frac{\partial f}{\partial x}(x_0,y_0)\) measures the rate of change of \(f\) at \((x_0,y_0)\) in the direction of the \(x\)-axis.
The partial derivative \(\frac{\partial f}{\partial y}(x_0,y_0)\) measures the rate of change of \(f\) at \((x_0,y_0)\) in the direction of the \(y\)-axis.

How do you compute partial derivatives?

Let’s talk about \(\displaystyle\frac{\partial f}{\partial x}(x,y)\) first. First, remember the general definition of the derivative: \[ f'(\mathbf{v})\mathbf{h} =\lim_{\lambda \to 0} \frac{f(\mathbf{v} + \lambda\mathbf{h}) - f(\mathbf{v})}{\lambda}. \]
Substituting \(\mathbf{v} = \begin{bmatrix} x \\ y \end{bmatrix}\) and \(\mathbf{h} = \mathbf{e}_1\) into the above definition, we get \[ \frac{\partial f}{\partial x}(x,y) = \lim_{\lambda \to 0} \frac{f(x+\lambda, y) - f(x,y)}{\lambda}. \]
Similarly, we have \[ \frac{\partial f}{\partial y}(x,y) = \lim_{\lambda \to 0} \frac{f(x, y+\lambda) - f(x,y)}{\lambda}. \]

How do you compute partial derivatives?

The above equations show that the partial derivatives are just single-variable derivatives with one of the two variables fixed! So:

To compute \(\displaystyle\frac{\partial f}{\partial x}(x,y)\), we treat \(y\) as a constant and differentiate \(f\) with respect to \(x\).
To compute \(\displaystyle\frac{\partial f}{\partial y}(x,y)\), we treat \(x\) as a constant and differentiate \(f\) with respect to \(y\).

Alternate notation

Sometimes we write \(f_x(x,y)\) instead of \(\displaystyle\frac{\partial f}{\partial x}(x,y)\).
Sometimes we write \(f_y(x,y)\) instead of \(\displaystyle\frac{\partial f}{\partial y}(x,y)\).

Exercise 1: Practice computing partial derivatives of functions \(\mathbb{R}^2 \to \mathbb{R}\)

Compute both partial derivatives of the following functions:

\(f(x,y) = 3x - x^2y^2+2x^3y\)
\(g(x,y) = xe^{x^2y}\)
\(h(x,y) = \ln(x^2+y^2)\)
\(k(x,y) = \sin(xy^2)\)

Partial derivatives and (total) derivatives, part 1

What is the relationship between the partial derivatives \[ \frac{\partial f}{\partial x}(x,y) \quad \text{and} \quad \frac{\partial f}{\partial y}(x,y) \] of a differentiable function \(f:\mathbb{R}^2\to\mathbb{R}\) and its (total) derivative \(f'(x,y)\)?
Remember that \[ \frac{\partial f}{\partial x}(x,y) = f'(x,y)\mathbf{e}_1 \quad \text{and} \quad \frac{\partial f}{\partial y}(x,y) = f'(x,y)\mathbf{e}_2. \]
Given a general step vector \(\mathbf{h} = \begin{bmatrix} h_1 \\ h_2 \end{bmatrix}\), we can write resolve it into the standard basis as \(\mathbf{h} = h_1\mathbf{e}_1 + h_2\mathbf{e}_2\).
Then, if \(f'(x,y) = \begin{bmatrix} a & b \end{bmatrix}\), we have \[ f'(x,y)\mathbf{h} = \begin{bmatrix} a & b \end{bmatrix} \begin{bmatrix} h_1 \\ h_2 \end{bmatrix} = ah_1 + bh_2. \]
However, we also have \[ f'(x,y)\mathbf{e}_1 = \begin{bmatrix} a & b \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = a \] and \[ f'(x,y)\mathbf{e}_2 = \begin{bmatrix} a & b \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = b. \]
Putting everything together, we get \[ f'(x,y)\mathbf{h} = \frac{\partial f}{\partial x}(x,y)h_1 + \frac{\partial f}{\partial y}(x,y)h_2. \]

Partial derivatives and (total) derivatives, part 2

Relation between partial derivatives and (total) derivatives (for \(\mathbb{R}^2 \to \mathbb{R}\))

Suppose that \(f: \mathbb{R}^2\to \mathbb{R}\) is a differentiable function. Then, as a \(1\times 2\) matrix, we have \[ f'(x,y) = \begin{bmatrix} \displaystyle\frac{\partial f}{\partial x}(x,y) & \displaystyle\frac{\partial f}{\partial y}(x,y) \end{bmatrix}. \]

Exercise 2: Practice computing (total) derivatives of functions \(\mathbb{R}^2 \to \mathbb{R}\)

Compute the (total) derivatives of the following functions:

\(f(x,y) = 3x - x^2y^2+2x^3y\)
\(g(x,y) = xe^{x^2y}\)
\(h(x,y) = \ln(x^2+y^2)\)
\(k(x,y) = \sin(xy^2)\)

Visualizing partial derivatives of functions \(\mathbb{R}^2 \to \mathbb{R}\)

In the very, very, very special case of functions of the form \(f: \mathbb{R}^2 \to \mathbb{R}\), we can draw the graph of \(f\) as a surface in \(\mathbb{R}^3\).
If \(f\) is differentiable at a point \((x_0,y_0)\) then its tangent plane is the graph of the function \[ \begin{align*} L(x,y) &= f'(x_0,y_0) \begin{bmatrix} x-x_0 \\ y-y_0 \end{bmatrix} + f(x_0,y_0) \\ &= \begin{bmatrix} \displaystyle\frac{\partial f}{\partial x}(x_0,y_0) & \displaystyle\frac{\partial f}{\partial y}(x_0,y_0) \end{bmatrix} \begin{bmatrix} x-x_0 \\ y-y_0 \end{bmatrix} + f(x_0,y_0) \\ &= \frac{\partial f}{\partial x}(x_0,y_0)(x-x_0) + \frac{\partial f}{\partial y}(x_0,y_0)(y-y_0) + f(x_0,y_0). \end{align*} \]
Thus, the slope of the tangent plane in the \(x\)-direction is \(\displaystyle\frac{\partial f}{\partial x}(x_0,y_0)\), while the slope in the \(y\)-direction is \(\displaystyle\frac{\partial f}{\partial y}(x_0,y_0)\).

Exercise 3: Plotting tangent planes of functions \(\mathbb{R}^2 \to \mathbb{R}\)

Compute and plot the tangent approximations and tangent planes of the following functions at the given points:

\(f(x,y) = 3x - x^2y^2+2x^3y\) at \((-1,0)\)
\(g(x,y) = xe^{x^2y}\) at \((-2,0)\)

Exercise 4: Plotting tangent lines of cross sections

Consider the function \(f:\mathbb{R}^2 \to \mathbb{R}\) defined by \(f(x,y)=x^2+y^2\). Compute and plot the tangent line to the cross section with \(x=1\), at the point \(y=1\). Interpret your answer visually.