Category Archives: Uncategorized

Homework for Wednesday 28 March

See the notes from Monday’s class here.

• Review the limits at infinity in the second part of those notes. Make sure that you understand how we see the limits at infinity and negative infinity in the graphs.

We will be discussing L’Hôpital’s Rule next time, which helps to find certain kinds of limits (including limits at infinity) which are otherwise hard or impossible to find. But you need to understand the basic limits, including the limits of basic functions as given in the notes, before you can use L’Hôpital’s Rule!

You may want to view this video from Khan Academy which is an introduction to L’Hôpital’s Rule (on YouTube, watch out for videos autoplaying after!)

• Finish the WeBWorK on Derivatives of Inverse Trig Functions (if you have not already done so)

 

Don’t forget, if you get stuck on a problem, you can post a question on Piazza. Make sure to give your question a good subject line and tell us the problem itself – we need this information in order to answer your question. And please only put one problem per posted question!

Notes for Monday 26 March class

(after Test 2)

Topics:

• Problem 7 from the WeBWorK assignment on Derivatives of inverse trig functions

My version of the problem: Find $\frac{\textrm{d}y}{\textrm{d}x}$ if

$y = \tan^{-1}(\sqrt{4x^{2} – 1})$

Recall that $\frac{\textrm{d}}{\textrm{d}x}\left(\tan^{-1}(x) = \frac{1}{1+x^{2}}$,

so

$\frac{\textrm{d}}{\textrm{d}u}\left(\tan^{-1}(u)\right) = \frac{1}{1+u^{2}}$

 

To find $\frac{\textrm{d}y}{\textrm{d}x}$ we need to use the Chain Rule:

$\frac{\textrm{d}y}{\textrm{d}x} = \frac{\textrm{d}}{\textrm{d}u}\left(\tan^{-1}(u)\right)\bigg|_{u = \sqrt{4x^{2} -1}}\cdot \frac{\textrm{d}}{\textrm{d}x}(\sqrt{4x^{2} – 1})$

 

We need to find the derivative of the inner function using the General Power Rule (special case of the Chain Rule):

$\frac{\textrm{d}}{\textrm{d}x}(\sqrt{4x^{2} – 1}) = \frac{\textrm{d}}{\textrm{d}x}\left[(4x^{2} – 1)^{\frac{1}{2}}\right]$

$= \frac{1}{2}(4x^{2} – 1)^{-\frac{1}{2}}\cdot 8x$

$= \frac{4x}{\sqrt{4x^{2} – 1}}$

 

Put this and the derivative of $\tan^{-1}(u)$ into the Chain Rule formula for $\frac{\textrm{d}y}{\textrm{d}x}$:

$\frac{\textrm{d}y}{\textrm{d}x} = \frac{\textrm{d}}{\textrm{d}u}\left(\tan^{-1}(u)\right)\bigg|_{u = \sqrt{4x^{2} -1}}\cdot \frac{\textrm{d}}{\textrm{d}x}(\sqrt{4x^{2} – 1})$

$= \frac{1}{1+u^{2}}\bigg|_{u = \sqrt{4x^{2} -1}}\cdot \frac{4x}{\sqrt{4x^{2} – 1}}$

$= \frac{1}{1+(\sqrt{4x^{2} -1})^{2}}\cdot \frac{4x}{\sqrt{4x^{2} – 1}}$

WeBWorK does not require that we simplify the answer any further than this (as long as we carefully type it in correctly! remember to use preview!) but this can be considerably simplified:

$ \frac{1}{1+(\sqrt{4x^{2} -1})^{2}}\cdot \frac{4x}{\sqrt{4x^{2} – 1}} = \frac{1}{1+ 4x^{2} -1}\cdot \frac{4x}{\sqrt{4x^{2} – 1}} = \frac{1}{4x^{2}}\cdot \frac{4x}{\sqrt{4x^{2} – 1}} = \frac{1}{x\sqrt{4x^{2} – 1}}$

So $\frac{\textrm{d}y}{\textrm{d}x} = \frac{1}{x\sqrt{4x^{2} – 1}}$

———————————————————————————————————————————-

Review of limits at infinity:

Remember that $\displaystyle \lim_{x\rightarrow \infty}f(x) = L$ means that as x grows without bound (goes off to the right), the height of the graph (value of f(x)) gets closer and closer to L,

and similarly $\displaystyle \lim_{x\rightarrow -\infty}f(x) = L$ means that as x goes in the negative direction without bound (goes off to the left), the height of the graph (value of f(x)) gets closer and closer to L.

If L is a number (not $\infty$ or $-\infty$), the fact that the limit exists means that there is a horizontal asymptote $y = L$ on that side.

$\displaystyle \lim_{x\rightarrow \infty}f(x) = \infty$ means that as x goes off to the right, the height of the graph (value of f(x)) grows without bound.

$\displaystyle \lim_{x\rightarrow \infty}f(x) = -\infty$ means that as x goes off to the right, the height of the graph (value of f(x)) goes in the negative direction without bound.

And similarly for limits as x approaches $-\infty$.

 

Here are some examples from basic functions you have already seen. Look at the graphs in Desmos which are linked here to see why these limits are what they are.

 

• $y = x^{2}$: graph

$ \lim_{x\rightarrow \infty}x^{2} = \infty$ and  $\lim_{x\rightarrow -\infty}x^{2} = \infty$

 

• $y = x^{3}$: graph

$ \lim_{x\rightarrow \infty}x^{3} = \infty$ and  $\lim_{x\rightarrow -\infty}x^{3} = -\infty$

 

• $y = e^{x}$: graph

$ \lim_{x\rightarrow \infty}e^{x} = \infty$ and  $\lim_{x\rightarrow -\infty}e^{x} = 0$.

$y = e^{x}$ has a horizontal asymptote on the left, which is $y=0$ (The x-axis)

 

• $y = \ln(x)$: graph

$ \lim_{x\rightarrow \infty}\ln(x) = \infty$ (hard to see, because the height grows very slowly! But if you keep going to the right the height will eventually be greater than any number your chooose.)   $\lim_{x\rightarrow -\infty}\ln(x)$ does not exist, because the domain of $\ln(x)$ is only the interval $(0, \infty)$.

 

• $y = \sin(x)$: graph

$ \lim_{x\rightarrow \infty}\sin(x)$ and  $\lim_{x\rightarrow -\infty}\sin(x)$ do not exist, because $\sin(x)$ oscillates between 1 and -1 through all the number in between as you go off either to the right or to the left.

 

• $y = \frac{1}{x}$: graph

$ \lim_{x\rightarrow \infty}\frac{1}{x} = 0$ and  $\lim_{x\rightarrow -\infty}\frac{1}{x} = 0$.

$y = \frac{1}{x}$ has a horizontal asymptote on both sides, which is $y=0$ (The x-axis)

 

 

Test 2 review UPDATED and with corrections!

Test 2 is rescheduled for the first hour or so of class on Monday 26 March.

MAT1475Test2ReviewSpring2018

A student found errors in the answers to 3a, b, and d. Here are the corrected answers:

MAT2572Test2ReviewAnswersSpring2018  (corrected)

 

Here are some resources related to these problems:

For problem 1, here is a video from PatrickJMT using the definition of the derivative to find the derivative

[link is to YouTube, so other videos may autoplay after]

For problem 3, here is a video from PatrickJMT showing many different derivatives using the rules of differentiation 

For problem 5, here are some examples of implicit differentiation from VisualMath [uses Flash]

Here is a video from PartickJMT on implicit differentiation

For problem 6, here is a video from PatrickJMT on logarithmic differentiation

 

Winter Storm Warning (again)

There is a winter storm warning for the region starting tonight through Thursday.

 

I recommend checking the college website to find out the status of classes. I, myself, cannot cancel class.

Also, sign up for CUNY alerts, but they seem to  run slow. Checking the college website or your City Tech email seems to be the best.

Homework for Wednesday 14 March

For notes from Monday’s class, see this post.

• Make sure that you complete the WeBWorK on the Chain Rule by Tuesday midnight! The assignment on Higher Derivatives is due Sunday midnight. But please don’t wait to the last minute!

• From the WeBWorK on Implicit Differentiation, do problem 6. (You may do more if you like, but do at least this one.)

There will be a Quiz on Wednesday. The quiz will be on using the Chain Rule to find the derivative of a function and then find the equation of a tangent line using that derivative.

 

Don’t forget, if you get stuck on a problem, you can post a question on Piazza. Make sure to give your question a good subject line and tell us the problem itself – we need this information in order to answer your question. And please only put one problem per posted question!

Monday 12 March class

Topics:

• More on the Chain Rule and the two special cases

Here is a revised version of the slideshow which contains the two special cases and also information about implicit differentiation:

MAT1475CompositeFnsChainRule-slideshow

• Examples using the General Power Rule:

$f(x) = \sqrt{4-x^{2}}$

$f(x) = \frac{1}{x^{2} + 5}$

• Examples using the Exponential Function Rule:

$f(x) = e^{3x}$ (was done last time)

$f(x) = e^{-x^{2}}$

• Using the Chain Rule along with the Product Rule

• Finding the equation of the tangent line to the graph of $y = \sqrt{4-x}$ at the point $(3,1)$

• Introduction to implicit differentiation: we assume y is a function of x and use the Chain Rule in the form $\frac{\textrm{d}}{\textrm{d}x}f(y) = \frac{\textrm{d}f}{\textrm{d}y}\frac{\textrm{d}y}{\textrm{d}x}$

Some examples:

$\frac{\textrm{d}}{\textrm{d}x}\left(y^{2}\right) = 2y\frac{\textrm{d}y}{\textrm{d}x}$

$\frac{\textrm{d}}{\textrm{d}x}\left(x^{3}y\right) = x^{3}\frac{\textrm{d}y}{\textrm{d}x} + 3x^{2}y$ (Use the Product Rule along with the Chain Rule.)

Then we found the slope of the tangent line to the graph $x^{2} + y^{2} = 16$, $y\ge 0$, at the point where $x = 1$.

Homework for Monday 12 March

For notes from Wednesday’s class, see this post.

• Make sure that you complete the WeBWorK on Product and Quotient Rules by Thursday midnight! (One-time extension). The assignment on Higher Derivatives is due Sunday midnight. But please don’t wait to the last minute! (Sorry, that assignment didn’t get posted in time.)

• From the WeBWorK on Chain Rule, do problems 1-5. (You may do more if you like, but do at least these.)

• If you have not already done so, make sure that you do the assigned problems from the textbook from last time.

There will be a Quiz on Monday. The quiz will be on the Chain Rule, a problem similar to the examples I worked in class (or problems 1-5 in the WeBWorK).

 

Don’t forget, if you get stuck on a problem, you can post a question on Piazza. Make sure to give your question a good subject line and tell us the problem itself – we need this information in order to answer your question. And please only put one problem per posted question!