During Class

Brain Gains

Definition: Partial Derivative

Consider a function of several variables $f(x,y)$.

The symbol $\frac{\partial f}{\partial x}$ is read "the partial derivative of $f$ with respect to $x$" or "the partial of $f$ with respect to $x$".
The symbol $\frac{\partial f}{\partial y}$ is read "the partial derivative of $f$ with respect to $y$" or "the partial of $f$ with respect to $y$".

To calculate a partial derivative we treat all independent variables, except the one we are differentiating with respect to, as constants and use the derivative rules for a single variable function.

Let's look at an example that includes the chain rule. Consider the function $S(x,y) = 3(\pi x + 4y)^3$.

The partial of $S$ with respect to $x$ is $\frac{\partial S}{\partial x} = 3(3(\pi x + 4y)^{2})(\pi)$.
And the partial of $S$ with respect to $y$ is $\frac{\partial S}{\partial y} = 3(3(\pi x + 4y)^{2})(4)$.

1. For the function $g(x) = e^{3x^2}$, identify two functions $f(u)$ and $u(x)$ so that $g(x) = f(u(x))$. Then compute $\frac{dg}{dx} = \frac{df}{du}\frac{du}{dx}$.

2. Use the information about $f$ and $f'$ in the table and $g$ to find each of the following values.

Given $h(x) = g(f(x))$, find $h'(3)$.
Given $k(x) = f(g(x))$, find $k'(1)$.

Solution

Each of these problems focus on using the chain rule. We will use $g(1) = 2$ and $g'(1)=3$ (slope of the line).

We have $h'(x) = g'(f(x))f'(x)$, which means $h'(3) = g'(f(3))f'(3) = g'(1)(6) = (3)(6) = 18$.
We get $k'(x) = f'(g(x))g'(x)$, and so $k'(1) = f'(g(1))g'(1) = f'(2)(3) = (5)(3) = 15$.

3. Given $f(a; \textbf{x}) = \sum_{i=1}^{3} (ax_i + 2)^5$, find $\frac{df}{da}$.

Answer

We see $\frac{df}{da} = 5x_1(ax_1 + 2)^4 + 5x_2(ax_2 + 2)^4 + 5x_3(ax_3 + 2)^4 = \sum_{i=1}^3 5x_i(ax_i + 2)^4$.

4. For the function $f(x) = e^{3x}$ and $g(x) = x^4 - 1$, write out both $f(g(x))$ and $g(f(x))$, and then compute the derivative of each composite function.

Solution

The composite functions are

$f(g(x)) = e^{3(x^4 - 1)}$ and
$g(f(x)) = (e^{3x})^4 - 1$.

The derivatives of $f$ and $g$ are $f'(x) = 3e^{3x}$ and $g'(x) = 4x^3$. The chain rule then gives us

$\ds\frac{d}{dx}[f(g(x))] = f'(g(x))g'(x) = 3e^{3(g(x))}4x^3 = 3e^{3(x^4 - 1)}4x^3.$
$\ds\frac{d}{dx}[g(f(x))] = g'(f(x))f'(x) = 4(f(x))^33e^{3x} = 4(e^{3x})^33e^{3x}.$

Group Meeting

Start by giving each person a moment to share what they chose to prepare for class. Help each other address any questions. When each person has had a chance to share, move on the other activities.

Activity - Chain Rule Practice

For each function $g(x)$, identify two functions $f(u)$ and $u(x)$ so that $g(x) = f(u(x))$ and then compute $\frac{dg}{dx} = \frac{df}{du}\frac{du}{dx}$.

$g(x) = (x^2-3)^4$
$g(x) = \ln(4x+3)$
$g(x) = e^{4x+3}$
$g(x) = (x^3+\ln x)^2$
$g(x) = \ln (e^{2x} + 1)$
$g(x) = e^{4\ln{5x}+3}$

Activity - Partial Derivatives with the Chain Rule

Given $S(x, y; a, b) = 3(ax + by)^{-3}$, find $\diff{S}{x}$ assuming all other variables and parameters are constant then find $\diff{S}{y}$ assuming all other variables and parameters are constant. (Yes, these are long-winded ways of saying compute $\frac{\partial S}{\partial x}$ and $\frac{\partial S}{\partial y}$.)
Given $g(a_1, a_2; \textbf{x},\textbf{y}) = \sum_{i=1}^{7} \sqrt{a_1x_i + a_2y_i}$, find $\frac{\partial g}{\partial a_1}$ and $\frac{\partial g}{\partial a_2}$.
Given $h(x, y; \textbf{a}, \textbf{b}) = \sum_{i=1}^{21} 3(a_ix + b_iy)^{-3}$, find $\frac{\partial h}{\partial x}$ and $\frac{\partial h}{\partial y}$.

Answer

We have $\frac{\partial S}{\partial x}(x, y; a, b) = -9a(ax + by)^{-4}$ and $\frac{\partial S}{\partial y}(x, y; a, b) = -9b(ax + by)^{-4}$.
We find $\frac{\partial g}{\partial a_1} = \sum_{i=1}^7 \frac{x_i}{2\sqrt{a_1x_i + a_2y_i}}$ and $\frac{\partial g}{\partial a_2} = \sum_{i=1}^7 \frac{y_i}{2\sqrt{a_1x_i + a_2y_i}}$.
We compute $\frac{\partial h}{\partial x} = \sum_{i=1}^{21} -9a_i(a_ix + b_iy)^{-4}$ and $\frac{\partial h}{\partial y} = \sum_{i=1}^{21} -9b_i(a_ix + b_iy)^{-4}$

Activity - Graphing a Second Derivative

For the function below, graph the first and second derivative.

Solution

See Desmos.

Activity - Loglikelihood

You have been given a list of 44 data points, $(t_i, y_i)$, were $t_i$ is the time in hours since the light was turned on and $y_i$ is the percent intensity of the bulb after $t_i$ hours. A model for the percent intensity of the bulb is $f(x;m,b) = mx+b$, where $t$ is the time in hours since the light was turned on.

Write the formula for the residuals.

Assume the residuals (errors) are independent and normally distributed (with mean of 0 and standard deviation of 1).

Write down the probability model we are using for one of the residuals.
What does knowing the residuals are independent allow us to do?
- Use the assumption of independent residuals to write the joint probability function.
Write the likelihood function for the residuals in this situation.
Write the loglikelihood function for the errors in this situation.
Rewrite the loglikelihood function using the properties of logs and sums.

Activity - Chain Rule Practice

For each pair of functions $f(x)$ and $g(x)$ below, write out the compositions $f(g(x))$ and $g(f(x))$. Then compute the derivative of each composite function. As a suggestion, compute both $f'(x)$ and $g'(x)$ first, and then show how you can combine $f,g,f',g'$ to get the derivative of each composite function. Remember to pass the chalk between each problem.

$f(x) = 3x + 4$, $g(x) = x^2 + 1$
$f(x) = e^x$, $g(x) = \sqrt{2x-5}$
$f(x) = \frac{1}{x}$, $g(x) = 2\ln(x)$
$f(x) = (x-5)^4$, $g(x) = \sqrt[3]{x^2}$
$f(x) = e^{3x}$, $g(x) = x^4 - 1$
$f(x) = \frac{x}{x-3}$, $g(x) = \ln(5x)$
$f(x) = 2x^3 - 10x^2 + 2x$, $g(x) = \sqrt{\frac{x-4}{2}}$
$f(x) = \frac{1}{x^5}$, $g(x) = 4e^x$
$f(x) = \sqrt{8-x}$, $g(x) = (2x-1)(x+5)$
$f(x) = e^{x/2}$, $g(x) = 5x^3$
$f(x) = 1 - \frac{3}{x^4}$, $g(x) = 2 - \sqrt{x}$
$f(x) = \frac{3}{x^2}$, $g(x) = \ln(x)$
$f(x) = 27 - x^3$, $g(x) = x^2 - x - 12$
$f(x) = \sqrt[3]{4-x^2}$, $g(x) = 5x + 7$
$f(x) = \frac{x+1}{x-2}$, $g(x) = e^{-x}$

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HomePage Schedule	Class 22 During Class Derivative Rules Brain Gains Definition: Partial Derivative Consider a function of several variables $f(x,y)$. The symbol $\frac{\partial f}{\partial x}$ is read "the partial derivative of $f$ with respect to $x$" or "the partial of $f$ with respect to $x$". The symbol $\frac{\partial f}{\partial y}$ is read "the partial derivative of $f$ with respect to $y$" or "the partial of $f$ with respect to $y$". To calculate a partial derivative we treat all independent variables, except the one we are differentiating with respect to, as constants and use the derivative rules for a single variable function. Let's look at an example that includes the chain rule. Consider the function $S(x,y) = 3(\pi x + 4y)^3$. The partial of $S$ with respect to $x$ is $\frac{\partial S}{\partial x} = 3(3(\pi x + 4y)^{2})(\pi)$. And the partial of $S$ with respect to $y$ is $\frac{\partial S}{\partial y} = 3(3(\pi x + 4y)^{2})(4)$. 1. For the function $g(x) = e^{3x^2}$, identify two functions $f(u)$ and $u(x)$ so that $g(x) = f(u(x))$. Then compute $\frac{dg}{dx} = \frac{df}{du}\frac{du}{dx}$. 2. Use the information about $f$ and $f'$ in the table and $g$ to find each of the following values. Given $h(x) = g(f(x))$, find $h'(3)$. Given $k(x) = f(g(x))$, find $k'(1)$. Solution Each of these problems focus on using the chain rule. We will use $g(1) = 2$ and $g'(1)=3$ (slope of the line). We have $h'(x) = g'(f(x))f'(x)$, which means $h'(3) = g'(f(3))f'(3) = g'(1)(6) = (3)(6) = 18$. We get $k'(x) = f'(g(x))g'(x)$, and so $k'(1) = f'(g(1))g'(1) = f'(2)(3) = (5)(3) = 15$. 3. Given $f(a; \textbf{x}) = \sum_{i=1}^{3} (ax_i + 2)^5$, find $\frac{df}{da}$. Answer We see $\frac{df}{da} = 5x_1(ax_1 + 2)^4 + 5x_2(ax_2 + 2)^4 + 5x_3(ax_3 + 2)^4 = \sum_{i=1}^3 5x_i(ax_i + 2)^4$. 4. For the function $f(x) = e^{3x}$ and $g(x) = x^4 - 1$, write out both $f(g(x))$ and $g(f(x))$, and then compute the derivative of each composite function. Solution The composite functions are $f(g(x)) = e^{3(x^4 - 1)}$ and $g(f(x)) = (e^{3x})^4 - 1$. The derivatives of $f$ and $g$ are $f'(x) = 3e^{3x}$ and $g'(x) = 4x^3$. The chain rule then gives us $\ds\frac{d}{dx}[f(g(x))] = f'(g(x))g'(x) = 3e^{3(g(x))}4x^3 = 3e^{3(x^4 - 1)}4x^3.$ $\ds\frac{d}{dx}[g(f(x))] = g'(f(x))f'(x) = 4(f(x))^33e^{3x} = 4(e^{3x})^33e^{3x}.$ Group Meeting Start by giving each person a moment to share what they chose to prepare for class. Help each other address any questions. When each person has had a chance to share, move on the other activities. Activity - Chain Rule Practice For each function $g(x)$, identify two functions $f(u)$ and $u(x)$ so that $g(x) = f(u(x))$ and then compute $\frac{dg}{dx} = \frac{df}{du}\frac{du}{dx}$. $g(x) = (x^2-3)^4$ $g(x) = \ln(4x+3)$ $g(x) = e^{4x+3}$ $g(x) = (x^3+\ln x)^2$ $g(x) = \ln (e^{2x} + 1)$ $g(x) = e^{4\ln{5x}+3}$ Activity - Partial Derivatives with the Chain Rule Given $S(x, y; a, b) = 3(ax + by)^{-3}$, find $\diff{S}{x}$ assuming all other variables and parameters are constant then find $\diff{S}{y}$ assuming all other variables and parameters are constant. (Yes, these are long-winded ways of saying compute $\frac{\partial S}{\partial x}$ and $\frac{\partial S}{\partial y}$.) Given $g(a_1, a_2; \textbf{x},\textbf{y}) = \sum_{i=1}^{7} \sqrt{a_1x_i + a_2y_i}$, find $\frac{\partial g}{\partial a_1}$ and $\frac{\partial g}{\partial a_2}$. Given $h(x, y; \textbf{a}, \textbf{b}) = \sum_{i=1}^{21} 3(a_ix + b_iy)^{-3}$, find $\frac{\partial h}{\partial x}$ and $\frac{\partial h}{\partial y}$. Answer We have $\frac{\partial S}{\partial x}(x, y; a, b) = -9a(ax + by)^{-4}$ and $\frac{\partial S}{\partial y}(x, y; a, b) = -9b(ax + by)^{-4}$. We find $\frac{\partial g}{\partial a_1} = \sum_{i=1}^7 \frac{x_i}{2\sqrt{a_1x_i + a_2y_i}}$ and $\frac{\partial g}{\partial a_2} = \sum_{i=1}^7 \frac{y_i}{2\sqrt{a_1x_i + a_2y_i}}$. We compute $\frac{\partial h}{\partial x} = \sum_{i=1}^{21} -9a_i(a_ix + b_iy)^{-4}$ and $\frac{\partial h}{\partial y} = \sum_{i=1}^{21} -9b_i(a_ix + b_iy)^{-4}$ Activity - Graphing a Second Derivative For the function below, graph the first and second derivative. Solution See Desmos. Activity - Loglikelihood You have been given a list of 44 data points, $(t_i, y_i)$, were $t_i$ is the time in hours since the light was turned on and $y_i$ is the percent intensity of the bulb after $t_i$ hours. A model for the percent intensity of the bulb is $f(x;m,b) = mx+b$, where $t$ is the time in hours since the light was turned on. Write the formula for the residuals. Assume the residuals (errors) are independent and normally distributed (with mean of 0 and standard deviation of 1). Write down the probability model we are using for one of the residuals. What does knowing the residuals are independent allow us to do? Use the assumption of independent residuals to write the joint probability function. Write the likelihood function for the residuals in this situation. Write the loglikelihood function for the errors in this situation. Rewrite the loglikelihood function using the properties of logs and sums. Activity - Chain Rule Practice For each pair of functions $f(x)$ and $g(x)$ below, write out the compositions $f(g(x))$ and $g(f(x))$. Then compute the derivative of each composite function. As a suggestion, compute both $f'(x)$ and $g'(x)$ first, and then show how you can combine $f,g,f',g'$ to get the derivative of each composite function. Remember to pass the chalk between each problem. $f(x) = 3x + 4$, $g(x) = x^2 + 1$ $f(x) = e^x$, $g(x) = \sqrt{2x-5}$ $f(x) = \frac{1}{x}$, $g(x) = 2\ln(x)$ $f(x) = (x-5)^4$, $g(x) = \sqrt[3]{x^2}$ $f(x) = e^{3x}$, $g(x) = x^4 - 1$ $f(x) = \frac{x}{x-3}$, $g(x) = \ln(5x)$ $f(x) = 2x^3 - 10x^2 + 2x$, $g(x) = \sqrt{\frac{x-4}{2}}$ $f(x) = \frac{1}{x^5}$, $g(x) = 4e^x$ $f(x) = \sqrt{8-x}$, $g(x) = (2x-1)(x+5)$ $f(x) = e^{x/2}$, $g(x) = 5x^3$ $f(x) = 1 - \frac{3}{x^4}$, $g(x) = 2 - \sqrt{x}$ $f(x) = \frac{3}{x^2}$, $g(x) = \ln(x)$ $f(x) = 27 - x^3$, $g(x) = x^2 - x - 12$ $f(x) = \sqrt[3]{4-x^2}$, $g(x) = 5x + 7$ $f(x) = \frac{x+1}{x-2}$, $g(x) = e^{-x}$ Edit Page History Source Attach File Backlinks List Group Page last modified on February 17, 2026, at 03:09 PM
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22

During Class

Brain Gains

Group Meeting

Activity - Chain Rule Practice

Activity - Partial Derivatives with the Chain Rule

Activity - Graphing a Second Derivative

Activity - Loglikelihood

Activity - Chain Rule Practice