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terenzreignzBest ResponseYou've already chosen the best response.1
Derivative of the sum is the sum of the derivatives.
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.1
Although, to be really creative, you can just recall that the derivative of a constant is 0 :D
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.1
sec²x = tan²x + 1
 one year ago

wioBest ResponseYou've already chosen the best response.1
\[ \sec^2x\tan^2x = \frac{1}{\cos^2x}  \frac{\sin^2 x}{\cos^2 x} \]
 one year ago

wioBest ResponseYou've already chosen the best response.1
So I mean you're gonna end up with y=1
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
\[\frac{d}{dx}(\sec^2x\tan^2x)=2\sec x(\sec x \tan x)2\tan x(\sec^2x) = 2\sec^2x tanx  2\sec^2x \tan x = 0 \]
 one year ago

jayshaneBest ResponseYou've already chosen the best response.0
the answer in the book is zero...
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
Please listen to @terenzreignz y = 1 Don't miss these sipmle ones.
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
Sorry it cut off a but, but the last one just cut off "tan x"
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.1
So many different ways to do it, but if all else eludes you, do it directly, as @kirbykirby But sec²x = tan²x + 1 So sec²x  tan²x = tan²x + 1  tan²x = 1 And the derivative of 1...?
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
Yes there are many ways to approach this one. You can rewrite sec^x and tan^x in terms of cos x and sin x, but if you know the derivative of sec x and tan x (I recommend that you t do), then it's easier to proceed that way
 one year ago

jayshaneBest ResponseYou've already chosen the best response.0
i will get the double prime if i will get the derivative 1...?
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.1
It might pay (and save you time) to know by heart your trigonometric identities, too.
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
It absolutely WILL pay huge dividends to brush up on your trigonometry. If you were taking my examine and you did ANYTHING on this problem except recognize IMMEDIATELY that \(\dfrac{d}{dt}[\sec^{2}(t)  \tan^{2}(t)\ = \dfrac{d}{dt}1 = 0\), I would consider recommending you for a refresher course in a lower level class. Gotta know your trig. It's not optional!
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
@jayshane I'm not sure what you mean about getting a double prime? But the derivative of 1 (or any constant) is just 0 :)
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
Ok don't get so uptight @tkhunny :S Oye. It's not THE end of the world to not see it immediately o_o. I;m past cal 3 and doing differential equations now and even I didn't see it right away as you mentioned it lool. And I mean I barely use trig anymore that I am doing stats now
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
I usually only think of these identities when I can't "plug and chug" right away lol
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
@kirbykirby No uptight going on here, but I do appreciate your calming influence. I feel very strongly about this particular issue. One should not be in differential equations without a very solid foundation in trigonometry. Simple as that. Also, as we have yet to see ANY effort by @jayshane, it's just not clear what we're doing, here. A little strong reality therapy seemed like it might be appropriate.
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
Well I don't think that's a fair assumption to make that I don't have a good foundation in trig because I didn't recognize that as THE first thing to do?! o_o
 one year ago

jayshaneBest ResponseYou've already chosen the best response.0
i acctually having a hard time dealing with trigonometric identities.. and im studying harder and harder on it its just im just an ordinary man.. but im really tring hard pls dont be anger on me...
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.1
Nobody's angry with you @jayshane But, if you have the time, do study trigonometry... :)
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
No we are not angry at you :) We were just recommending to really know the trig identities as much as you can because they save you time :)
 one year ago

jayshaneBest ResponseYou've already chosen the best response.0
ive been memorizing it but i dont know how to use them in solving..
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
@ jayshane No anger, just encouragement. Get a trig book and work all 4,000 problems. Get it on your head and it will pay great dividends. @kirbykirby seems to be a good case that it isn't absolutely necessary, but if you could spend ALL your time in Differential Equations studying Differential Equations, don't you think this would be better than spending 1/2 your time in Differential Equations trying to remember trigonometry?
 one year ago

jayshaneBest ResponseYou've already chosen the best response.0
ok thank you for the answers i will study more...
 one year ago

jayshaneBest ResponseYou've already chosen the best response.0
but for now please help me just last 2 questions..!!
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
Good call. Show your work, too. This will help us as much as it helps you.
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
Well I am not spending 1/2 my time remembering trig :P I do remember the identities but I just don't necessarily apply them right away unless I am stuck, or just force myself to be "oh maybe I should use a trig identity here". Do you know what I mean?
 one year ago

jayshaneBest ResponseYou've already chosen the best response.0
my teacher encourages us to use the long method and i just cant get it,,
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
And anyway even if you do forget, it's very easy to pull up a cal book and look for the identities, (although you should know them for the exam!!)
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
@kirbykirby Right. No ridicule intended. "1/2" was a bit hyperbolic. There are many survival methodologies. I was kind of a freak in my first differential equations class, so that does tend to make me think my survival method is the only way to go. I do realize there are humans on this planet. :)
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.1
@jayshane Your teacher recommends the long method for your practice It's not on every opportunity that knowledge of trigonometric identities will save you from long calculations, BUT the long method is the only method that would work for all situations... Now, you said you have two more questions?
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
@tkhunny Ok good to know we are on good terms :P As for @jayshane, you still need help right
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
@jayshane I can think of only one reason to encourage a class to use the "long method". Your teacher knows your differential calculus isn't very good and is forcing you to practice! :)
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.1
Chain rule, @jayshane A good way to work on this is to identify what you could call the "outermost" function, the function that sort of "contains" all the others. Which function seems to fit this description?
 one year ago

jayshaneBest ResponseYou've already chosen the best response.0
she does just give this assignment to us and she doesnt even give explanations she just gave the identities and were having a hard time to deal with it..!
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.1
Well, @jayshane That's why we're here :) Now, which function is outermost?
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.1
Ahh, but sin itself is inside tangent, no?
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
Yeah A good trick for the chain rule as @terenzreignz mentions, identity your outer functions and inner functions for the chain rule. Think of what the elementary function is and what changed: Here, we could think of tan x being the elementary function, but we changed x to "xsin x". So, you can think of it like a box: you have xsinx in the box tan[ ] , so you find the derivative of tan [ ] and multiply by the derivative inside the box [ ]
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.1
y=tan(xsinx) And I've already named the outermost function... So you know tan contains all others, what's the derivative of tan?
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.1
leave out the du/dx, and stick to sec² x So "write that down" except instead of x, copy whatever was "inside" the tan function, namely, xsin x (NOTE: We're not yet done!)
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
\(f(x) = \tan(x\cdot\sin(x))\)  What a fascinating function. I'm not sure I've ever seen that one in a problem set.
 one year ago

oldrin.batakuBest ResponseYou've already chosen the best response.0
Remember the Pythagorean identity? \(\cos^2x+\sin^2x=1\); if you divide through by \(\cos^2x\), it yields \(1+\tan^2x=\sec^2x\). We can apply this identity to simplify our function! $$y=\sec^2x\tan^2x=\tan^2x+1\tan^2x=1$$Do you know what the derivative of a constant function is? What is the slope of a horizontal line?
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.1
Well, I like to think of chain rule as sort of a hitman's list, and think of it this way, you've already "taken 'tan' out of the picture" so you go for its "inner circle" lol anyway So now that you've "dealt with" tan (lol) time to deal with whatever was inside it. So remember sec²(xsin x) ? You multiply this to whatever the derivative of the inner function is, namely, xsin x. So what's the derivative of xsin x?
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
@oldrin.bataku We have covered the answer for that already :) We are approaching a new question. Sorry to burst your bubble
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.1
No... remember xsin x is the product of two functions, now would be a great time to recall your product rule?
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
That\s not exactly right. You have x*sin x, not just "sin x" In fact you have a product rule with functions "x" and "sin x" being multiplied
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
d/dx (x*sinx) = (1)sinx + x(cosx) by the product rule
 one year ago

oldrin.batakuBest ResponseYou've already chosen the best response.0
@kirbykirby I posted to explain the identity because none of you had. @jayshane Why don't we break our function into pieces? and *chain* the derivatives together? $$u=x\sin x\\y=\tan v\\\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=(\sec^2u)(\sin x+x\cos x)=\sec^2(x\sin x)[\sin x+x \cos x]$$
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.1
@oldrin.bataku I think I did post the identity, though I didn't derive it. @jayshane I can see where you're getting at, for as long as dsin x is the derivative of sin x and dx is the derivative of x. What's the derivative of sin x? What's the derivative of x?
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.1
Also, you don't multiply them, you add them.
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.1
Hang on, let me rewrite your statement properly, in a way that not even your teacher can find an error in :) \[x\frac{d}{dx}\sin x +(\sin x)\frac{d}{dx}x\] Now work out these derivatives.
 one year ago

oldrin.batakuBest ResponseYou've already chosen the best response.0
@jayshane the product rule tells us \(\frac{d}{dx}uv=\frac{du}{dx}v+\frac{dv}{dx}u\)  see how we multiply one of the original functions by the derivative of the other?.
 one year ago

oldrin.batakuBest ResponseYou've already chosen the best response.0
In our case of \(\frac{d}{dx}x\sin x\), we have two functions being multiplied: \(x\) and \(\sin x\). Do you know how to determine the derivative of this product using the product rule?
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
I actually find it easier to forget the "u" notation when first learning the chain rule. Just look at it as outer and inner functions. And you won't ever need the unotation once you master it that way
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.1
@kirbykirby This is product rule now, though... Eventually we do do away with the u and v, probably
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
Oops I meant the product rule LOL sorry. I am responding to another post and I am getting confused between both posts
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.1
Almost... What's the derivative of sin x?
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
No not quite. d/dx (sin x) = cos x d/dx (cos x ) = sin x... yes it's confusing at first
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.1
Yeah, derivative of sine is cosine Proof on request.
 one year ago

jayshaneBest ResponseYou've already chosen the best response.0
can u give me appropriate solution guys, then ill just study it..
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.1
Well, if you accept that the derivative of sin x is cos x, then you can work out (PLEASE DO WORK IT OUT) That the derivative of xsin x is x(cos x) + sin x Which as a whole expression, you multiply that one we got earlier, which was sec²(xsin x) Hence, you get [sec²(xsin x)][x(cos x) + sin x]
 one year ago

jayshaneBest ResponseYou've already chosen the best response.0
then how will i continue to solve? the answer in the book is 4tanxsec2x
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
Does your prof require that form of an answer (4tanx sec2x)?
 one year ago

jayshaneBest ResponseYou've already chosen the best response.0
not exactly it is just an answer in the book.
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
Because I can guarantee you it's the answer :) http://www.wolframalpha.com/input/?i=d%2Fdx+tan%28xsinx%29
 one year ago

jayshaneBest ResponseYou've already chosen the best response.0
another question sir kirbs Y=sec4xtan4x
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
I tried equation your textbook answer with that one and I don't get an equality... which means the book might have a typo. Either way, it is way too much work to get to an answer like that :)
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
You can proceed in a similar way to what I did. Or, you can make use of the identity: 1+tan^2 x = sec^2 x So:
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
\[\sec^4x\tan^4x = (\sec^2x)^2\tan^4x\] can you see that?
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
\[=(1+\tan^2x)^2\tan^4x\] \[=(1+2\tan^2x+\tan^4x)\tan^4x\] by using the identity (a+b)^2 = (a+2ab+b^2)
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
\[=1+2\tan^2x\]now our problem is much simpler! We find \[\frac{d}{dx}(1+2\tan^2x)=\frac{d}{dx}1+\frac{d}{dx}2\tan^2x\]
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
\[=0+2\frac{d}{dx}\tan^2x\] \[=2*(2*\tan x)(\sec^2 x) \] since this is the famous chain rule again. You put down the ^2 from tan( ) in front, and you multiply by the derivative of tan x which is sec^2 x
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
\[=4\tan x \sec^2x\]
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
Think of your box: you have \[[\tan x]^2 = 2[\tan x]*\frac{d}{dx}[\tan x]\]
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
In general for a "power" you have like \[[f(x)]^n=n[f(x)]^{n1}*\frac{d}{dx}f(x)\] I hope I am not confusing you by writing this
 one year ago

jayshaneBest ResponseYou've already chosen the best response.0
=(1+2tan2x+tan4x)−tan4x how did u get this kirb?
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
There should be a derivative d/dx in front of [f(x)]^n, sorry my computer was lagging
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
Ok you know that \[(a+b)^2 = (a+2ab+b^2)\]
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
\[(1+\tan^2x)^2=(1+2(1)(\tan^2x)+(\tan^2x)^2) \].... here we have that \[a = 1, b = \tan^2x\]
 one year ago

jayshaneBest ResponseYou've already chosen the best response.0
THANK YOU A LOT KIRB...
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.2
Using the identity \[1+\tan^2x=\sec^2x\]We have \[\sec^2x\tan^2x=(1+\tan^2x)\tan^2x\] \[=1\] So, \[\frac{d}{dx}1=0\]
 one year ago
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