Turn in #2 is due Tuesday, Sept. 27, 2016!
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Hey guys. Anyone know the difference between a corner and a cusp. Had some discussion about it but never really came to a consensus. Thoughts?
ReplyDelete-Jacob Edelson
A cusp has slopes approaching infinite values from either direction (ie. the curve is approaching verticality) while a corner may not have such behavior (although, you could construct a piece-wise function where approaching from one direction the slope is becoming very large and from the other direction the slope is a nice finite value and this would technically be a corner). There is no real value in distinguishing between corners and cusps.
DeleteQuestion from SP pg 111 #17-22: should we be figuring these out by graphing them and looking to see where they are differentiable, or is there an algebraic method we should be using (for example in problem 17 you can identify two vertical asymptotes where it is not differentiate but in 18 it's hard to just know where it won't work without using a calculator).
ReplyDeleteThe idea of these problems is to use any method you can to identify points where the function is not differentiable, so graphing each is fine.
DeleteThank you!
DeleteWhen taking the derivative from an equation a large amount times, what is the fastest way to get to the final answer?
ReplyDeleteIf you can't use a calculator, the fastest way is just deriving it one time at a time.
Delete- Andrew Saad
But if the derivative is greater than any of the exponents,and the exponents are positive integers, then you can just remove those terms.
Delete- Andrew Saad
It depends on the equation and what number derivative you're taking. If the number derivative that you're taking is greater than the number of most of the exponents, then it's worth it because then you just need to find the factorial of the exponent multiplied by the leading coefficient. But, if very few terms "max out" then it'll take more time to do that process. Also note that the "max out" number is actually one greater than the exponent's number because it'll become a constant before is goes to zero.
ReplyDelete-Jacob Edelson
For the first of the AP-style problems that were handed out today, how would we find what b had to be? I understood how to find a.. the derivatives of the two equations had to be equal, but confused on how to find b and why it matters if it will disappear after the derivative is taken anyway.
ReplyDelete-Jacob Edelson
Differentiability implies continuity...
Deleteyou're probably over thinking. If you know a, solve algebraically based on the fact that the function should be continuous.
ReplyDeleteWhat is an example of a critical point other than where f'(x)=0? Are non differentiable points all considered critical?
ReplyDelete~Sam Zerafa
There is not another example of a critical point because the definition of a critical point states that x=c is a critical point of the function f(x) if f(c) exists and if f'(c)=0 OR f'(c) does not exist. Because of this definition, there will not be another example of a critical point other than at f'(x)=0 because it can either be this or it can not exist. I don't know if all non differentiable points are considered critical but I do know that all points that do not exist are critical points but I'm not sure how the two connect.
DeleteCritical points are X values at which a function's derivative is equal to zero or undefined. Absolute value functions have critical points, but never have a slope of zero.
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ReplyDeleteWhat approach should are you all taking for #2 on the turn-in? I got to the step where you factor out an e^x in the numerator while taking the derivative but I'm not sure where to go from here.
ReplyDeleteYou have to think about it in order to get e^x alone- the value e is just a number (2.7 something) and if h is getting smaller and smaller than it is getting closer and closer to zero. In the step you said you're at, e is to the power of h inside the parenthesis. So if you think of it, any number to the (almost zero) is what? That may have been confusing but I don't want to straight out tell you the answer. Try plugging in a really small number for h (like .0000003) and see what happens.
DeleteRemember that the limit of a product of two functions is the product of their limits. (Try separating the limit that you have into two separate limits).
Delete-Matt H.
I used f(x+h)-f(x)/h and plugged x+h in where it should go and solved that to prove that e^x equals e^x.
DeleteHow should I approach #5 on the turn in? I've tried a few things, but I want to see if i'm on the right track?
ReplyDeletePower rule will be most useful.
DeleteShould we explain in words the parts of number five or is just plugging in and showing with number enough?
ReplyDeleteIf a problems says "explain", then a brief statement is necessary in addition to the work. If a problem says "show", then no words are needed, sufficient work is enough.
DeleteAny hints on number 1c of the AP-style questions on the turn in? I was thinking of trying to find an equation to represent the derivative of g(x), but I think that is not the correct direction to head in.
ReplyDeleteChoose the smallest subinterval containing the given x value, then look at what you know about the change occuring on this subinterval.
DeleteFor the chain rule worksheet, do you only take the derivative of what comes before the parentheses if what's in the parentheses is attached to a sin, cos, etc.?
ReplyDeleteuse the product rule, and then anything left in parentheses use the chain rule on. So, no to only with sin, cos, etc.
Delete- Andrew Saad
Im not really sure how to approach 4d on the turn in. Any hints?
ReplyDeleteThere's one theorem that makes the problem a lot more manageable. If I recall correctly, you should be able to find it in Chapter 4 of the textbook.
DeleteYou cannot use a theorem we have not learned/proved yet. There are several ways of showing this answer using what we already know.
DeleteFor number 4d, is there some kind of equation or theorem that could help figure out the answer?
ReplyDeletePerhaps IVT for derivatives applies here...or not...just a thought :)
DeleteOn p.140 #22, is the product rule necessary as we take the second derivative?
ReplyDeleteYes, as you will obtain a product of two functions when taking the derivative of cscx.
DeleteOn p.146 #4, how long do you carry out the chain rule for? I got to y=sec^2(2x-x^3)X(2-3x^2). Do I have the take the derivative of the last parentheses also?
ReplyDeleteThe last quantity of 2-3x^2 IS a derivative, so taking the derivative of it would be inappropriate. Look at the original function and determine what is the inner most function in the composition. Once you take the derivative of this inner most function, you are done.
DeleteFor #11 on the Implicit Differentiation Worksheet I'm getting stuck after finding dy/dx. Looking at the answer key I'm not sure what to do after that because plugging in 2 gives you an answer with variables in it.
ReplyDeleteHow would you do #11 for curve B on the packet we got on Friday? I found the derivative of the function but I'm not sure where to go from there.
ReplyDelete