Stony Brook MAT 125 Spring 2015
Lecture 15: Review for second midterm
March 30, 2015

Start   Ah okay there wont be any limits on the exam I dont know why i put this up but i did notice i made a correction there was another correction in here which somebody spotted, i forgot who it was These are from last year Number 7 i think it was find the derivative of sin inverse e^x
0:32the solution... see how it says 1 over 1 minus e^x squared?
the square root is missing that should be the square root of 1 over 1 minus e^x squared so number 7, ill fix that i havent had a chance to go on blackboard so as i said before, you should be able to take the derivative of anything
1:00and you should be able to use the derivative so lets do some practice problems Dan, you have a suggestion for the class?
do you want to teach the whole class?
so lets see... y=sin(t)-cos(t) and t equals...
1:39so find dy/dx thats not so hard, you guys can do that. just do a little substituting first so first thing you wanna do youre going to want to substitute the t
2:00so youre just going to want to make this sin of 1+cos^2x minus cos(1+cos^2x) and you guys can do it from there and its just chain rule. just chain rule
3:16so i'll do this, and you guys can copy so dy/dx would be what?
well of course you have to do the derivative of sin (1+cos^2x) Dan thought of the problem so its his fault
3:30just so you know actually its from my book the derivative of cos(1+cos^2x) times the derivative of (1+cos^2x) which is... the derivative of 1 is 0 2 cosx to the 1, negative sinx now were gonna go to the cosine part
4:01which is.., minus negative sin of 1+cos^2x and again times the derivative of 1+cos^2x which is.. 2cosx(-sinx) and you can clean that up a lot but you will not be required to simplify on the exam you simplify when you need to simplify
if you want to find zeros if you want to find the horizontal tangent increasing decreasing, that kind of stuff then simplifying will help you out but you dont have to simplify your answer alright, speaking of tangent lines... find the equation
5:04of the line
5:48dont worry about that okay?
find the equation tangent line to sin^2x=x^2+y^2- (pi^2)/16 at (pi/4, 0)
6:10are we ready?
do we need another minute or are we ready?
were ready. alright so first.. think of sin^2(x) as sin(x)^2 so now we need implicit differentiation because we have our x's and y's grouped together so we want to do the derivative since the derivative of sin^2(x) is 2sinx to the 1 times the derivative of sinx which is cos x
6:40this equals 2x plus 2y(dy/dx) and the derivative of that annoying pi/16 is just 0 we dont care about that okay now we have to plug in pi/4 for x and 0 for y oo..that didnt come out well
7:06okay, so were going to have a problem right here because we get nothing for dy/dx of course you all know what sin of pi/4 is its radical 2 over 2 and cos pi/4 is also radical 2 over 2 so you put them together, you get 1/2 so this becomes 2 times a half
7:30you see what our problem was?
so theres no dy/dx okay so lets do another one this time i know it'll work because i wont come up with crazy numbers
8:30okay, now itll work so you guys know what to do, i just added another term so well do this together, since we just did it so i threw in the two y terms so now if you take the derivative you get 2x+2y (dy/dx)-2(dy/dx) equals the derivative of sin^2x
9:00and the derivative of sin^2x is 2sinx to the 1 cosx now we plug in pi/4 and you get... 2(pi/4) + 0 - 2(dy/dx) = 2cos(pi/4)sin(pi/4)
9:42this is pi/2 - 2(dy/dx) = 2....radical 2...radical 2... times 2 radical 2 over 2
10:08so you get pi/2 - 2(dy/dx) = 1 thats not so bad so pi/2 - 1 = 2(dy/dx) divide it by 2 you get pi/4 - (1/2) = dy/dx
10:36the equation is... y-0= (pi/4) -(1/2) times x-(pi/4) wasnt that something?
11:05so where does 2(dy/dx) come from?
the derivative of x^2 is 2x the derivative of y^2 is 2y(dy/dx) and the derivative of 2y is 2(dy/dx) because any time you take the derivative of a y term you have to multiply it by dy/dx student inaudibly asks a question its 2y(dy/dx) (student continues to ask question)
11:37the derivative of y^3 is 3y^2 (dy/dx) okay?
in fact, youre going to see on monday okay?
remember.. when we do the derivative of x^2, its 2x(dx/dx) but dx/dx is 1 so you dont need to write that but if we were doing dx/dt
12:02you would get 2x (dx/dt) you have a term there okay?
student asks question this? we plug in 0 for y, so this becomes 0 okay?
student asks question ah, well, whats the square root of 2 times the square root of 2?
2 times 2 is four, divided by 4 is 1 okay?
12:38well first of all you take the derivative itll get you some credit then you plug in numbers and solve dy/dx that gets you some credit and then you write the equation of the line that gets you some credit okay?
so thats how you get partial credit on this and i will weight it heavily towards the first part alright
13:00how about those logarithmic differentiation thingys we had yesterday alright lets figure out what to erase im going to erase this okay y= x + e^x all raised to the sinx find dy/dx
13:35how do i know its going to be logarithmic differentiation?
notice how the function is raised to another function okay?
sure alright thats long enough so we have to use logarithmic differentiation here theres five steps to logarithmic differentiation
14:03always the same five, same order first thing you do... take the log of both sides natural log of y = the natural log of this okay so thats step 1 step one, take the ln of both sides step 2
14:31okay bring down the power and thats part of our log rules our log laws that you remember from last semester okay?
so take the log of both sides bring the power in front step 3 take the derivative of both sides
15:02and the derivative of the left side would just be a simple derivative of the log the derivative of the right side would require the product rule so the left side is going to be 1/y (dy/dx) its always 1/y (dy/dx) and you take the derivative of y the right side you need the power rule so sinx times the derivative of log of x+e^x so anytime you have the derivative of a log of a function
15:31you take the function and put it in the bottom of the fraction you take the derivative, and put it in the top of the fraction function on the bottom, derivative on the top plus other way around take the derivative of sin, which is cosine and leave the other term alone log of x+e^x okay so thats the crucial step
16:00take the log of both sides you bring the power in front and then you take the derivative and gotta use the product rule on the right sides now you just have to multiply both sides by y you get dy/dx y times this thing sin of this 1+e^x
16:31x+e^x e plus cosine x log of x + e^x okay?
and then you replace that with the original function you put y back okay?
which im not in the mood to right another time so youre just going to replace this
17:03with (x+e^x)^sinx if you forget to do that on the exam i have no idea if we're going to take off for that or not we'll discuss that tomorrow student inaudibly asks a question yeah i put the brackets... well, its y times this whole expression okay?
so you can use parentheses, you dont have to use square brackets but you have to use something to demonstrate you know its y times a whole thing
17:31and then you want to replace y with (x+e^x)^sinx okay?
should we do another one of those?
yes, wow
18:03okay find the derivative so, step one take the log of both sides ln(y) ln(1+tanx)^ (1-tanx)
18:34nice and easy step step 2 take the derivative-no bring the power in front 1-tanx times the log of (1+tanx) dont forget the log okay take the derivative 1/y (dy/dx) =
19:04gotta use the product rule so we can take the first term 1-tanx times the derivative of the second term 1+tanx in the denominator sec^2x in the numerator plus... log of 1+tanx
19:33times the derivative of 1-tanx which is minus sec^2x okay?
how'd we do on this so far?
good, bad, medium?
okay put y back with y times 1-tanx[(sec^2x)/(1+tanx)]
20:08plus... ln(1+tanx) minus sec^2x and then its too far down to write take y and replace it with (1+tanx)^(1-tanx) how'd we do on that?
huh? better?
yeah you got it, alright
20:32how about one of those annoying tangent line through the point things that we were doing yesterday okay..
21:51the tangent line to... y=x^2+2 passes through the point (2,0) find the point (a,b)
22:01which would be on the curve where the tangent line intersects the curve alright so we did a couple of these yesterday so lets think about whats going on here
22:42we have the curve y=x^2+2 and then the point (2,0) so the tangent line... is intersecting the curve somewhere up there at (a,b) okay?
23:00and the tangent line is going through (2,0) so remember the important thing about tangent lines theres 2 things to remember 1 is at the point of intersection, the slope of the tangent line is the same as the slope of the curve and the other thing is that the point of intersection is the same for both the tangent line and the curve so that point lies on the tangent line and on the curve so if we know its (a,b)
23:30we know how to find b if we know what a is because you just take a and plug it into the equation okay if a is 5 then b would be 5 squared plus 2 if a was 10...b would be 10 squared plus 2 thats why we have equations once you have x you can find y so this is really the point (a, a^2+2) okay so now what do we know well like i said we know that the slope is the same to the curve and to the tangent line
24:01so what are the slope of a line? well its the difference in y's over the difference in x's so the slope... is (a^2+2 - 0)/(a-2) difference in y's over difference in x's but what else do we know about the slope?
well the slope is the derivative so whats the derivative of y=x^2+2
24:32its just 2x so at (a,b) the slope is what?
2a okay?
so the slope has to equal 2x now we can solve the equation for a if you cross multiply you get.. a^2+2=2a(a-2)
25:03so a^2+2=2a^2-4a okay?
mistake? no oh im sure it doesnt factor in order to make it work well i have to actually think about the numbers so you can simplify this you get a^2-4a-2=0
25:30and then you do the quadratic formula a=4 plus/minus the square root of 16+8 over 2 theres actually 2 points where this works the point here on the right and theres some point something times that okay?
and i dont really care what the y coordinate is so i would say find coordinate b, yeah thats sin(b) so you make your numbers work out nicely
that takes an extra couple minutes of work you have the concept should we do another one of these?
i'll wait until everyone is done talking then we'll do another one so this time we have a curve y=8-x^2 which looks something like that
26:31tangent line... something like that and thats our point (a,b) also known as (a, 8-a^2) so the slope.. is 8-a^2
27:01minus 0 over a-6 but we know the derivative is -2x so this has to equal... -2a so cross multiply you get... 8-a^2= -2a(a-6)
27:33or 8-a^2= -2a^2 +12a put all the terms on one side you get a^2-12a+8=0 i told you i can make the numbers work sort of oops forgot a formula, sorry
28:00almost made the numbers work so a is 12+/- square root of 144-32 all over 2, and then whatever okay?
i didnt hear you, what?
you cant fit 2 out of 12, no if you wanted to simplify this.. its 12+/- the square root of 112 over 2
28:36which is 12 +/ 2 radical 28 4 radical 7 over 2 which is 6+/- 2 radical 7 6+ 2 radical 7, 6- 2 radical 7
29:00so you would take either one and you plug it back into the y=8-x^2 for the y value we'll give you an easy number if we give you something like this maybe there wont be anything like this on the exam and this was just a fun 20 minutes you learned some cool stuff very important to learn cool stuff but if we were to give you one for some crazy reason then the numbers will come out okay i hope alright, how about another thing
29:33we ready?
can i erase?
30:41suppose i have f(x) so y=x^3-12x^2+36x-10 find where the function is decreasing okay these are not very hard you'll get the hang of it
31:01how do you find where its increasing or decreasing well we've been doing a bunch of this stuff very similar but after this, wednesday we have class dont forget we've got all the tools and now we're going to start learning to apply them to a variety of problems okay?
so very exciting make sure you keep all the derivative tools so one of the big ones we're going to do is where functions are increasing and decreasing and where they get maximums and minimums
31:30maximum, minimum so take the derivative you get 3x^2-36x+36 i mean its 24x okay?
set that equal to 0 so 3x^2-24x+36=0
32:00factor out the 3 you get x^2 - 8x +12 =0 that factors nicely you get (x-2) and (x-6) so, at this point you have found where the derivative is 0 we need to figure out where the derivative is negative so when you want to do things like that
32:30its always the same technique youre going to sign test s-i-g-n sign test so you need to figure out what is the sign of the derivative in different parts of the number line okay?
so you take a number left of 2 a number between 2 and 6 and a number greater than 6 and you plug it into the derivative and you see if you get positive values or negative values so you take 0 for example thats a nice number
33:00you plug it in here you get 0-2 which is a negative number 0-6 is a negative number and a negative times a negative gets you a positive so the function is going to be increasing everywhere less than 2 then you take a number between 2 and 6 like 3 3-2 is a positive number 3-6 is a negative number a positive times a negative is negative you can plug it into any of these derivatives they're all the same thing
33:30its just easier when you plug it into the factored form now you take a number greater than 6 like 7 7-2 is a positive number 7-6 is a positive number a positive times a positive is positive so its increasing when you're less than 2 or greater than 6 so its decreasing when x is between 6 and 2 or informal notation (2,6) okay?
34:01should i give you one more?
we done for the day or should we do one more?
i hear a lot of one mores do you want another increasing/decreasing?
i'll give you a fast one we'll do it together rather than waiting suppose we had something really evil like...
34:35y=sin^2x and we want to figure out where is that increasing or decreasing so i would take the derivative of dy/dx (2sinx) to the 1 (cosx) and i need to figure out where is that 0
35:00on the interval 0 to 2pi well lets see... its going to be 0 where sin is 0 or its going to be 0 where cosine is 0 so that would be at x=0, pi/2, pi, 3pi/2 remember trigonometry is fair game on this test supposed to know your trig alrighty well anywhere either cosine or sin =0
35:31sin is 0 at 0 and pi cosine is 0 at pi/2 and 3pi/2 you're not counting 2pi okay, and let see on the plot where its increasing so now lets start plugging in numbers so lets try pi/4 thats 2, sin(pi/4) is positive
36:00cos (pi/4) is positive positive lets do 3pi/4 sin (3pi/4) is positive, were in the second quadrant cosine is negative so this is negative-its going to alternate by the way and 5pi/4 is positive 7pi/4 is negative so its increasing, decreasing, increasing, decreasing if you were to graph it, you'd see it okay?
alright, thats enough for today
36:31see you guys tomorrow, good luck