<p>Okay I really need some help with calc.</p>
<p>This one was from the quiz and I just want to verify my answer.</p>
<p>What is the derivative of the cubed root of x-2? Perform the calculation by hand and on the calculator and compare your answers?</p>
<p>I got 0 by hand and 100 on the calculator and said the calculator was off. Is that the correct way by hand?</p>
<p>Then I have a vague question from the quiz too that I remember. I couldn't determine if something was a cusp or corner. The function was either a square root or cubed root of some absolute value function. Could that be either..or depending on what function or are roots of absolute value functions always one or the other. Does it matter if cubed or square because I forget which it was? Sorry if that's confusing, but I need to know bad because I think I did poorly.</p>
<p>These are from homework:</p>
<p>Find der. of tan x/ (1 + sin x). I have no idea how to do this. I know that the derv. of tan is sec squared and sin is cos, but no clue how to do this.</p>
<p>Find the x values where the tangent lines of the tan function (from 0 to 2pi) are parallel to the line y=4x-7. I think I should set sec sq. equal to 4 but I don't know how to do that and what to do afterwards.</p>
<p>Ok, so that's 4 questions..sorry, but help is appreciated.</p>
<p>SORRY one more I forgot to post from the quiz that I want to check. It showed a picture of a hyperbola coming from (-3,-1) down to (-1, -4) and a line going from (0,2) to (3,4) and said that this was the derivative of a function. It then said to make a continuous graph with f(-1)=-1 of the ORIGINAL FUNCTION.</p>
<p>Sorry to bomb you guys with these, but I appreciate it because my teacher doesn't teach</p>
<p>How could you get 0 and 100 when you weren't given an x-value? Or did you simply forget to write that value down on here?</p>
<p>For the derivative of tanx/(1+sinx), you should use the quotient rule.</p>
<p>0 and 100. oh yeah, i forgot, it said at x=2 I used the calculator NDERV ( cubed root x-2,x,2) and it got 100 but by hand I got 0, so I need that cleared up.</p>
<p>I did the quotient and got (sec^2x tan^2x-tanx cos x)/(1+sinx)^2 is that right or should I do more or something else?</p>
<p>logisticswizard, I'll report this thread to the colleges you're applying to!</p>
<p>[(-1/3)((x-2)^(1/3))] would be the answer for the first question as written, using the Chain Rule.
Edit: Then, substituting in 0 one would get 0 as the final answer.</p>
<p>what a dork... i didnt even know a calc can do derivatives...hahaha</p>
<p>yeah, we learn chain rule tomorrow so i didn't know that but I got 0 anyways. so that quiz question is right.</p>
<p>what about the others?</p>
<p>come on gang... i'll try to help you with your homework if you help me with these other problems of mine. Especially check the trig der. i posted, the y=4x-7 one and the cusp/corner.</p>
<p>Find der. of tan x/ (1 + sin x) = tanx * (1+sinx)^-1
d/dx tanx * ( 1+sinx)^-1 + d/dx(1+sinx)^-1 * tan x<br>
to do d/dx(1+sinx)^-1 let u = 1+ sinx and du=cosx
so d/dx u^-1 = -1u^(-1-1) du = -cosx/(1+sinx)^2
so d/dx tanx * ( 1+sinx)^-1 + -cosx/(1+sinx)^2 * sinx/cos x
and the second one = -sinx/(1+sinx)^2</p>
<p>also you can use (lodehi - hidelo)lo^2 where hi=top one, lo = bot, de = d/dx</p>
<p>Find the x values where the tangent lines of the tan function (from 0 to 2pi) are parallel to the line y=4x-7. I think I should set sec sq. equal to 4 but I don't know how to do that and what to do afterwards.
4= 1/cos^2 (x)
4 cos^2 x = 1 so basicaly cos x= root 1/4 and these roots can be positive or negative
so cosx = 1/2 which is an x 2x xroot3 triangle and it is 30degrees. So you have 30 degrees = pi/6 and then you can also have 5pi/6 and also 7pi/6 and also 11pi/12 becase its cos^2 = 1/4 so the signs dont matter, they cancel.</p>
<p>Lucky 24 aint on til later. I think they are right, atleast the process should be.</p>
<p>wow! that's confusing..is that chain rule? we never do anything that complex in class and he just gives it for homework no explanation...okay.</p>
<p>so now just the parallel line and cusp/corner question please :)
oh and now I have a VERY strong feeling that my test is not gonna have be very high. it'll probably be in the 70s or low 80s, at least the people I sit by are usally in the 50s and 60s so I wont feel so bad when he hands it back. luckily i confused him with BS work so he wont take off a ton of points. hopefully he just makes each of our 20 questions worth 2 points instead of 5 and take off only 1 if we show work. then i'd have in the 90s. (that is not unheard of either, he did that in pre-calc because everyone would have had 20s and 30s if he didn't, everyone ended up with 70s-90s, talk about nice curve!)</p>
<p>okay thanks...can anyone verify those quick and still the cusp ones.</p>
<p>it helps matt, appreciate it greatly.</p>
<p>say that you have like d/dx of sinx/x thats the same thing let lo=x hi= sinx dehi=cosx so (lodehi-hidelo)/lo^2 = (xcosxdx - sinxdx)/x^2 which you can also do by the chain rule, which is what I did.
The derivative of 1/x is -1dx/x^2. because you take 1 away from the power, and its already a -1 so it becomes a -2 which is a +2 on the bottom. so like d/dx 1/x^2 = d/dx X^-2 = -2X^(-2-1) = -2/X^3. Hope this way is more readable.</p>
<p>Yep, I understand the high and low part. I just did the high and low and got this big abomination (sec^2 x+ sin x)(1+sinx)/(1+sinx)^2.</p>
<p>First I got cos x tanx + sec^2 x (1+sin x) / lo^2</p>
<p>Then I made that cos/1 sin/cos and the cos crossed out leaving me sinx+sec^2 x (1+sin x)/(1+sin x)^2</p>
<p>Maybe it would be easiest if you could tell me what to do from this point? or what I did wrong with the above.</p>
<p>tan x/ (1 + sin x)
((1+sinx) sec^2x - tanx(cosx)) / (1+sinx)^2 = the answer. It can be simplified slightly</p>
<p>sec^2x / (1+sinx) - sinx/(1+sinx)^2 if you really call that simplifying. It can take on many forms but these are probably the likely ones.</p>
<p>okay, thanks a bunch!!</p>
<p>real quick---the cusp/corner question anyone?</p>
<p>Absolute value functions have corners. The roots of such functions have cusps at their zeros (even functions only).</p>
<p>and odd functions would have nothing or corners? sorry to be a pain</p>
<p>bump sorry, just that one last ?^^^^</p>