Multivariable calculus: Exercises

10 Partial derivatives, part 2

Exercise 1: Computing partial derivatives of functions

Given a differentiable function \(f:\mathbb{R}^n \to \mathbb{R}\), recall that you can compute the partial derivatives

\[ \frac{\partial f}{\partial x_1}(x_1,x_2,\ldots,x_n), \quad \frac{\partial f}{\partial x_2}(x_1,x_2,\ldots,x_n), \quad \ldots, \quad \frac{\partial f}{\partial x_n}(x_1,x_2,\ldots,x_n) \]

by holding one variable constant and differentiating the other, just like in single-variable calculus.

Compute the partial derivatives of the following functions:

  1. \(f(x,y,z) = \sqrt{x^2+y^2+z^2}\)
  2. \(f(x,y,z) =\displaystyle\frac{3xz}{x+y}\)
  3. \(f(x,y,z) = \sin{(x+2y+3z)}\)

  1. Answers: \[ \begin{align*} \frac{\partial f}{\partial x}(x,y,z) &= \frac{x}{\sqrt{x^2+y^2+z^2}}, \\ \frac{\partial f}{\partial y}(x,y,z) &= \frac{y}{\sqrt{x^2+y^2+z^2}}, \\ \frac{\partial f}{\partial z}(x,y,z) &= \frac{z}{\sqrt{x^2+y^2+z^2}}. \end{align*} \]

  2. Answers: \[ \begin{align*} \frac{\partial f}{\partial x}(x,y,z) &= \frac{3yz}{(x+y)^2}, \\ \frac{\partial f}{\partial y}(x,y,z) &= -\frac{3xz}{(x+y)^2}, \\ \frac{\partial f}{\partial z}(x,y,z) &= \frac{3x}{x+y}. \end{align*} \]

  3. Answers: \[ \begin{align*} \frac{\partial f}{\partial x}(x,y,z) &= \cos{(x+2y+3z)}, \\ \frac{\partial f}{\partial y}(x,y,z) &= 2\cos{(x+2y+3z)}, \\ \frac{\partial f}{\partial z}(x,y,z) &= 3\cos{(x+2y+3z)}. \end{align*} \]

Exercise 2: Computing higher-order partial derivatives of functions

Remember that the second-order partial derivatives of a function are the partial derivatives of the first-order partial derivatives. Similarly, the third-order partial derivatives are the partial derivatives of the second-order partial derivatives, and so on.

Compute all third-order partial derivatives of the function \[ f: \mathbb{R}^2 \to \mathbb{R}, \quad f(x,y) = x^2y^3 + 3xy + 5y^2. \]

First-order:

\[ \begin{align*} \frac{\partial f}{\partial x}(x,y) &= 2xy^3 + 3y, \\ \frac{\partial f}{\partial y}(x,y) &= 3x^2y^2 + 3x + 10y. \end{align*} \]

Second-order:

\[ \begin{align*} \frac{\partial^2 f}{\partial x^2}(x,y) &= 2y^3, \\ \frac{\partial^2 f}{\partial y \partial x}(x,y) &= \frac{\partial^2 f}{\partial x \partial y}(x,y) = 6xy^2 + 3, \\ \frac{\partial^2 f}{\partial y^2}(x,y) &= 6x^2y + 10. \end{align*} \]

Third-order:

\[ \begin{align*} \frac{\partial^3 f}{\partial x^3}(x,y) &= 0, \\ \frac{\partial^3 f}{\partial y \partial x^2}(x,y) &= \frac{\partial^3 f}{\partial x^2 \partial y}(x,y) = \frac{\partial^3f}{\partial x \partial y \partial x}(x,y) = 6y^2, \\ \frac{\partial^3 f}{\partial y^2 \partial x}(x,y) &= \frac{\partial^3 f}{\partial x \partial y^2}(x,y) = \frac{\partial^3 f}{\partial y \partial x \partial y}(x,y) = 12xy, \\ \frac{\partial^3 f}{\partial y^3}(x,y) &= 6x^2. \end{align*} \]

Exercise 3: Computing total derivatives of functions

If \(f:\mathbb{R}^n\to \mathbb{R}\) is a differentiable function, then as we saw in class, the (total) derivative of \(f\) at a point \((x_1, x_2, \ldots, x_n)\) is given by the formula

\[ f'(x_1, x_2, \ldots, x_n) = \begin{bmatrix} \displaystyle\frac{\partial f}{\partial x_1}(x_1, x_2, \ldots, x_n) & \displaystyle\frac{\partial f}{\partial x_2}(x_1, x_2, \ldots, x_n) & \cdots & \displaystyle\frac{\partial f}{\partial x_n}(x_1, x_2, \ldots, x_n) \end{bmatrix}. \]

Compute the total derivatives of each of the functions in the first exercise:

  1. \(f(x,y,z) = \sqrt{x^2+y^2+z^2}\)
  2. \(f(x,y,z) =\displaystyle\frac{3xz}{x+y}\)
  3. \(f(x,y,z) = \sin{(x+2y+3z)}\)

  1. \(f'(x,y,z) = \begin{bmatrix} \displaystyle\frac{x}{\sqrt{x^2+y^2+z^2}} & \displaystyle\frac{y}{\sqrt{x^2+y^2+z^2}} & \displaystyle\frac{z}{\sqrt{x^2+y^2+z^2}} \end{bmatrix}\)

  2. \(f'(x,y,z) = \begin{bmatrix} \displaystyle\frac{3yz}{(x+y)^2} & -\displaystyle\frac{3xz}{(x+y)^2} & \displaystyle\frac{3x}{x+y} \end{bmatrix}\)

  3. \(f'(x,y,z) = \begin{bmatrix} \cos{(x+2y+3z)} & 2\cos{(x+2y+3z)} & 3\cos{(x+2y+3z)} \end{bmatrix}\)

Exercise 4: Concavity and partial derivatives

Let \(f: \mathbb{R}^2\to \mathbb{R}\) be a differentiable function. Recall from class that the partial derivative \(\frac{\partial f}{\partial x}(x_0,y_0)\) can be thought of as the slope of the tangent line at \(x=x_0\) to the cross section of the graph of \(f\) with the plane \(y=y_0\). And similarly for \(\frac{\partial f}{\partial y}(x_0,y_0)\). Use this idea, along with your knowledge about the link between concavity and second-derivatives from single-variable calculus, to answer the following questions.

  1. Is the graph of the function \(z=f(x,y) = x^2-y^2\) concave up or concave down in the \(x\)-direction at the point \((0,0)\)? How about in the \(y\)-direction?
  2. Is the graph of the function \(z=f(x,y) = 3xy^2-2y + 5x^2y^2\) concave up or concave down in the \(x\)-direction at the point \((1,1)\)? How about in the \(y\)-direction?
  3. I give you a differentiable function \(f:\mathbb{R}^2\to \mathbb{R}\), a fixed point \((x_0,y_0)\), and I tell you that \[ \frac{\partial f}{\partial x}(x_0,y_0) = \frac{\partial f}{\partial y}(x_0,y_0) = 0 \] and \[ \frac{\partial^2 f}{\partial x^2}(x_0,y_0) > 0, \quad \frac{\partial^2 f}{\partial y^2}(x_0,y_0) > 0. \] Tell me what you might believe about the graph of \(f\) at the point \((x_0,y_0)\).

  1. In the \(x\)-direction, we have \[ \frac{\partial^2 f}{\partial x^2}(0,0) = 2 > 0, \] so the graph is concave up in the \(x\)-direction at \((0,0)\). In the \(y\)-direction, we have \[ \frac{\partial^2 f}{\partial y^2}(0,0) = -2 < 0, \] so the graph is concave down in the \(y\)-direction at \((0,0)\).

  2. In the \(x\)-direction, we have \[ \frac{\partial^2 f}{\partial x^2}(1,1) = 10 > 0, \] so the graph is concave up in the \(x\)-direction at \((1,1)\). In the \(y\)-direction, we have \[ \frac{\partial^2 f}{\partial y^2}(1,1) = 16 > 0, \] so the graph is concave up in the \(y\)-direction at \((1,1)\).

  3. Since the first-order partial derivatives are both zero, you know that the tangent plane to the graph of \(f\) at \((x_0,y_0)\) is horizontal. Since the second-order partial derivatives are both positive, you also know that the graph of \(f\) is concave up in both the \(x\)-direction and the \(y\)-direction at \((x_0,y_0)\). So you might believe that the graph of \(f\) has a local minimum at \((x_0,y_0)\). Congratulations if you got this far. However, you’d be wrong! It’s not enough to check the concavity in the \(x\)- and \(y\)-directions, you need to check the concavity in all directions. For that, however, you need the so-called Hessian matrix, which we’ll get to later.