During Class

Brain Gains

1. Compute $k'(x)$ for $k(x) = (4x^3-7x)^5$.

Solution

We can let $y(u) = u^5$ and $u(x) = 4x^3-7x$. The then compute $$ \begin{align*} k'(x) &= \frac{dy}{du}\frac{du}{dx} &\text{(chain rule)} \\ &= (5u^4)(12x^2-7) &\text{(power rule)}\\ &= (5(4x^3-7x)^4)(12x^2-7) &\text{(write in terms of $x$)}. \end{align*} $$

2. Compute $g'(x)$ for $g(x) = (e^x+3x^2)(x^3-2\ln x)$.

Solution

Let $u(x) = e^x+3x^2$ and compute $u'(x) = e^x+6x$ using the sum, constant multiple, power, and exponential rules. Then let $v(x) = x^3-2\ln x$ and obtain $v'(x) = 3x^2-\frac{2}{x}$, using the power, difference, and logarithm rules. The product rule then gives $$\begin{align*} \frac{d}{dx}[(e^x+3x^2)(x^3-2\ln x)] &=\frac{d}{dx}[u(x)v(x)] \\ &=u'(x)v(x)+ u(x)v'(x)\\ &=(e^x+6x)(x^3-2\ln x)+ (e^x+3x^2)(3x^2-\frac{2}{x}). \end{align*}$$ We won't simplify it any past this. Our goal is to just practice using the rules at this point.

3. Compute $h'(x)$ for $h(x) = \dfrac{e^x+3x^2}{x^3-2\ln x}$.

Solution

Let $u(x) = e^x+3x^2$ and compute $u'(x) = e^x+6x$ using the sum, constant multiple, power, and exponential rules. Then let $v(x) = x^3-2\ln x$ and obtain $v'(x) = 3x^2-\frac{2}{x}$, using the power, difference, and logarithm rules. The quotient rule then gives $$\begin{align*} \frac{d}{dx}\left(\dfrac{e^x+3x^2}{x^3-2\ln x}\right) &=\frac{d}{dx}\left(\dfrac{u(x)}{v(x)}\right) \\ &=\left(\dfrac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\right) \\ &=\left(\dfrac{(e^x+6x)(x^3-2\ln x) - (e^x+3x^2)(3x^2-\frac{2}{x})}{(x^3-2\ln x)^2}\right). \end{align*}$$ We won't simplify it any past this. Our goal is to just practice using the rules at this point.

4. For the function $f(x) = x^4$, let $g(x)=f'(x)$. Compute $g'(x)$. Then compute $D_x(g'(x))$. Finish by computing $\frac{d}{dx}(D_xg')$.

Answers

$ g(x) = f'(x) = 4x^3 $
$ g'(x) = 12x^2 $. We often write this as $f''(x)$ or $\frac{d^2f}{dx^2}$.
$ D_x(g'(x)) = 24x $. We often write this as $f'''(x)$ or $\frac{d^3f}{dx^3}$.
$ \frac{d}{dx}(D_xg'(x)) = 24 $. We often write this as $f^{ (4) }(x)$ or $\frac{d^4f}{dx^4}$. Once we pass 3 derivatives, prime notation changes - who wants to count the number of primes.

5. Give an equation of the linear approximation to $f(x)=x^3$ at $x=2$. (What point does the line pass through? What is the slope of the line?)

Solution

We know $f'(x) = 3x^2$, and so the slope of this line is $f'(2) = 12$. The linearization (tangent line) passes through the point $(2,f(2))=(2,8)$. This gives an equation of the linearlization as $$y-8=12(x-2).$$ We can solve for $y$ to obtain $$y=8+12(x-2) \quad\text{or}\quad y=12x-16.$$ All three of the forms above are correct. We can plot the function along with it's linearization in Desmos.

We can also plot the function along with it's linear approximation (tangent line) in R.

x <- seq(-2,4,0.001)
plot(x,x^3,type='l',ylim=c(-5,15),ylab="y")
lines(x,8+12*(x-2),col=4)
points(2,8,pch=16,col=2)
abline(h=0,col='gray',lty=3)
abline(v=0,col='gray',lty=3)

6. What is the late penalty for not getting your project work in by the Due Date?

Solution

There isn't one. The point to the deadline is to help you keep up, and provide you time for revisions as needed.

Once the "End Date" of each Task/Project passes, then no more submissions will be accepted. These end dates are listed in the syllabus.

If you got stuck on something, it's OK. There is time to ask questions. We'll take time in class periodically to help each other.

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 - Compute derivatives

Derivative Rules

Take turns at the chalkboard computing the derivative of each function below. Show your work, step-by-step, listing the rules you use at each stage. Remember to pass the chalk after each problem.

$f(x) = (3x-2x^2)(5+4x)$. Remember to list each rule you used. Then pass the chalk (PTC)
$f(x) = \frac{x-1}{2x+3}$ (PTC)
$f(x) = \left( \frac{1}{x} + 1 \right)(x-1)$ (PTC)
$f(t) = (2t^2 - 3)(4 - t^2 - t^4)$ (PTC - I'll stop writing this.)
$f(p) = (p^3 - 2)^2$
$f(x) = \frac{2x^2 - 4x + 3}{2 - 3x}$
$f(x) = \frac{7}{x^3}$
$f(x) = \frac{4x^2 - 3x}{8\sqrt{x}}$
$f(x) = \left( \frac{x-3}{x+4} \right)(x^2 + 2x +1)$
$f(x) = \frac{ (1 - 2x)(3x+2) }{5x-4}$
$f(x) = 5 x^3 e^x \ln x$

Final Answers

These are the final answers you should obtain after listing out each step.

$\diff{f}{x}(x) = (3-4x)(5+4x) + (3x - 2x^2)(4)$
$\diff{f}{x}(x) = \frac{ (1)(2x+3) - (x-1)(2) }{ (2x+3)^2 }$
$\diff{f}{x}(x) = (-x^{-2})(x-1) + (x^{-1} + 1)(1)$
$\diff{f}{t}(t) = (4t)(4 - t^2 - t^4) + (2t^2 - 3)(-2t-4t^3)$
$\diff{f}{p}(p) = (3p^2)(p^3-2) + (p^3-2)(3p^2)$
$\diff{f}{x}(x) = \frac{ (4x - 4)(2 - 3x) - (2x^2 - 4x + 3)(-3) }{ (2 - 3x)^2 }$
$\diff{f}{x}(x) = \frac{ (0)(x^3) - (7)(3x^2) }{ (x^3)^2 }$
$\diff{f}{x}(x) = \frac{ (8x - 3)(8\sqrt{x}) - (4x^2 - 3x)(4x^{-1/2}) }{ (8\sqrt{x})^2 }$
$\diff{f}{x}(x) = \frac{ (1)(x+4) - (x-3)(1) }{ (x+4)^2 }(x^2 + 2x +1) + \left( \frac{x-3}{x+4} \right)(2x + 2)$
$\diff{f}{x}(x) = \frac{ ((-2)(3x+2) + (1-2x)(3))(5x-4) - ((1-2x)(3x+3))(5) }{ (5x-4)^2 }$
Try searching online for a tool that will compute derivatives for you. Use it to check your work on this one.

Activity - Higher Order Derivatives

Since the derivative of a function is a function, we can find the derivative of a derivative. In fact, we can continue to take derivatives to find the third derivative, fourth derivative, etc. When we take a derivative of a derivative we find a higher order derivative. The second derivative $f$ we can denote with $$f''(x) = \frac{d}{dx}\left(\frac{d}{dx}(f)\right) = \frac{d^2f}{dx^2} = D_x^2 f.$$ We use similar notation for higher order derivatives, such at $f'''(x)$ for the third derivative and $\frac{d^4f}{dx^4}$ for a 4th derivative. We avoid prime notation and use notation such as $f^{ (7) }(x)$ when the number of derivatives is quite large.

Open this Higher Order Derivatives worksheet and take turns computing derivatives, passing the chalk after each problem.

Bonus Problems

Suppose we know that $f_i(x)$ is a differentiable function and $c_i$ is a constant for each integer $i$ from 1 to 4.

Compute $\frac{d}{dx}\left(\sum_{i=1}^{4}c_if_i(x)\right)$.
Compute $\frac{d}{dx}\left(\prod_{i=1}^{4}c_if_i(x)\right)$.
Let $n$ be a positive integer, and then repeat both of the above after replacing 4 with $n$.

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HomePage Schedule	Class 19 During Class Derivative Rules Brain Gains 1. Compute $k'(x)$ for $k(x) = (4x^3-7x)^5$. Solution We can let $y(u) = u^5$ and $u(x) = 4x^3-7x$. The then compute $$ \begin{align} k'(x) &= \frac{dy}{du}\frac{du}{dx} &\text{(chain rule)} \\ &= (5u^4)(12x^2-7) &\text{(power rule)}\\ &= (5(4x^3-7x)^4)(12x^2-7) &\text{(write in terms of $x$)}. \end{align} $$ 2. Compute $g'(x)$ for $g(x) = (e^x+3x^2)(x^3-2\ln x)$. Solution Let $u(x) = e^x+3x^2$ and compute $u'(x) = e^x+6x$ using the sum, constant multiple, power, and exponential rules. Then let $v(x) = x^3-2\ln x$ and obtain $v'(x) = 3x^2-\frac{2}{x}$, using the power, difference, and logarithm rules. The product rule then gives $$\begin{align} \frac{d}{dx}[(e^x+3x^2)(x^3-2\ln x)] &=\frac{d}{dx}[u(x)v(x)] \\ &=u'(x)v(x)+ u(x)v'(x)\\ &=(e^x+6x)(x^3-2\ln x)+ (e^x+3x^2)(3x^2-\frac{2}{x}). \end{align}$$ We won't simplify it any past this. Our goal is to just practice using the rules at this point. 3. Compute $h'(x)$ for $h(x) = \dfrac{e^x+3x^2}{x^3-2\ln x}$. Solution Let $u(x) = e^x+3x^2$ and compute $u'(x) = e^x+6x$ using the sum, constant multiple, power, and exponential rules. Then let $v(x) = x^3-2\ln x$ and obtain $v'(x) = 3x^2-\frac{2}{x}$, using the power, difference, and logarithm rules. The quotient rule then gives $$\begin{align} \frac{d}{dx}\left(\dfrac{e^x+3x^2}{x^3-2\ln x}\right) &=\frac{d}{dx}\left(\dfrac{u(x)}{v(x)}\right) \\ &=\left(\dfrac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\right) \\ &=\left(\dfrac{(e^x+6x)(x^3-2\ln x) - (e^x+3x^2)(3x^2-\frac{2}{x})}{(x^3-2\ln x)^2}\right). \end{align}$$ We won't simplify it any past this. Our goal is to just practice using the rules at this point. 4. For the function $f(x) = x^4$, let $g(x)=f'(x)$. Compute $g'(x)$. Then compute $D_x(g'(x))$. Finish by computing $\frac{d}{dx}(D_xg')$. Answers $ g(x) = f'(x) = 4x^3 $ $ g'(x) = 12x^2 $. We often write this as $f''(x)$ or $\frac{d^2f}{dx^2}$. $ D_x(g'(x)) = 24x $. We often write this as $f'''(x)$ or $\frac{d^3f}{dx^3}$. $ \frac{d}{dx}(D_xg'(x)) = 24 $. We often write this as $f^{ (4) }(x)$ or $\frac{d^4f}{dx^4}$. Once we pass 3 derivatives, prime notation changes - who wants to count the number of primes. 5. Give an equation of the linear approximation to $f(x)=x^3$ at $x=2$. (What point does the line pass through? What is the slope of the line?) Solution We know $f'(x) = 3x^2$, and so the slope of this line is $f'(2) = 12$. The linearization (tangent line) passes through the point $(2,f(2))=(2,8)$. This gives an equation of the linearlization as $$y-8=12(x-2).$$ We can solve for $y$ to obtain $$y=8+12(x-2) \quad\text{or}\quad y=12x-16.$$ All three of the forms above are correct. We can plot the function along with it's linearization in Desmos. We can also plot the function along with it's linear approximation (tangent line) in R. x <- seq(-2,4,0.001) plot(x,x^3,type='l',ylim=c(-5,15),ylab="y") lines(x,8+12(x-2),col=4) points(2,8,pch=16,col=2) abline(h=0,col='gray',lty=3) abline(v=0,col='gray',lty=3) 6. What is the late penalty for not getting your project work in by the Due Date? Solution There isn't one. The point to the deadline is to help you keep up, and provide you time for revisions as needed. Once the "End Date" of each Task/Project passes, then no more submissions will be accepted. These end dates are listed in the syllabus. If you got stuck on something, it's OK. There is time to ask questions. We'll take time in class periodically to help each other. 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 - Compute derivatives Derivative Rules Take turns at the chalkboard computing the derivative of each function below. Show your work, step-by-step, listing the rules you use at each stage. Remember to pass the chalk after each problem. $f(x) = (3x-2x^2)(5+4x)$. Remember to list each rule you used. Then pass the chalk (PTC) $f(x) = \frac{x-1}{2x+3}$ (PTC) $f(x) = \left( \frac{1}{x} + 1 \right)(x-1)$ (PTC) $f(t) = (2t^2 - 3)(4 - t^2 - t^4)$ (PTC - I'll stop writing this.) $f(p) = (p^3 - 2)^2$ $f(x) = \frac{2x^2 - 4x + 3}{2 - 3x}$ $f(x) = \frac{7}{x^3}$ $f(x) = \frac{4x^2 - 3x}{8\sqrt{x}}$ $f(x) = \left( \frac{x-3}{x+4} \right)(x^2 + 2x +1)$ $f(x) = \frac{ (1 - 2x)(3x+2) }{5x-4}$ $f(x) = 5 x^3 e^x \ln x$ Final Answers These are the final answers you should obtain after listing out each step. $\diff{f}{x}(x) = (3-4x)(5+4x) + (3x - 2x^2)(4)$ $\diff{f}{x}(x) = \frac{ (1)(2x+3) - (x-1)(2) }{ (2x+3)^2 }$ $\diff{f}{x}(x) = (-x^{-2})(x-1) + (x^{-1} + 1)(1)$ $\diff{f}{t}(t) = (4t)(4 - t^2 - t^4) + (2t^2 - 3)(-2t-4t^3)$ $\diff{f}{p}(p) = (3p^2)(p^3-2) + (p^3-2)(3p^2)$ $\diff{f}{x}(x) = \frac{ (4x - 4)(2 - 3x) - (2x^2 - 4x + 3)(-3) }{ (2 - 3x)^2 }$ $\diff{f}{x}(x) = \frac{ (0)(x^3) - (7)(3x^2) }{ (x^3)^2 }$ $\diff{f}{x}(x) = \frac{ (8x - 3)(8\sqrt{x}) - (4x^2 - 3x)(4x^{-1/2}) }{ (8\sqrt{x})^2 }$ $\diff{f}{x}(x) = \frac{ (1)(x+4) - (x-3)(1) }{ (x+4)^2 }(x^2 + 2x +1) + \left( \frac{x-3}{x+4} \right)(2x + 2)$ $\diff{f}{x}(x) = \frac{ ((-2)(3x+2) + (1-2x)(3))(5x-4) - ((1-2x)(3x+3))(5) }{ (5x-4)^2 }$ Try searching online for a tool that will compute derivatives for you. Use it to check your work on this one. Activity - Higher Order Derivatives Since the derivative of a function is a function, we can find the derivative of a derivative. In fact, we can continue to take derivatives to find the third derivative, fourth derivative, etc. When we take a derivative of a derivative we find a higher order derivative. The second derivative $f$ we can denote with $$f''(x) = \frac{d}{dx}\left(\frac{d}{dx}(f)\right) = \frac{d^2f}{dx^2} = D_x^2 f.$$ We use similar notation for higher order derivatives, such at $f'''(x)$ for the third derivative and $\frac{d^4f}{dx^4}$ for a 4th derivative. We avoid prime notation and use notation such as $f^{ (7) }(x)$ when the number of derivatives is quite large. Open this Higher Order Derivatives worksheet and take turns computing derivatives, passing the chalk after each problem. Bonus Problems Suppose we know that $f_i(x)$ is a differentiable function and $c_i$ is a constant for each integer $i$ from 1 to 4. Compute $\frac{d}{dx}\left(\sum_{i=1}^{4}c_if_i(x)\right)$. Compute $\frac{d}{dx}\left(\prod_{i=1}^{4}c_if_i(x)\right)$. Let $n$ be a positive integer, and then repeat both of the above after replacing 4 with $n$. Edit Page History Source Attach File Backlinks List Group Page last modified on February 10, 2026, at 09:55 AM*
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19

During Class

Brain Gains

Group Meeting

Activity - Compute derivatives

Activity - Higher Order Derivatives

Bonus Problems