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anonymous
 3 years ago
how about differentiate: y=sec²xtan²x
anonymous
 3 years ago
how about differentiate: y=sec²xtan²x

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terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1Derivative of the sum is the sum of the derivatives.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ \sec^2x\tan^2x = \frac{1}{\cos^2x}  \frac{\sin^2 x}{\cos^2 x} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So I mean you're gonna end up with y=1

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the answer in the book is zero...

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.0Please listen to @terenzreignz y = 1 Don't miss these sipmle ones.

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

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1So 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...?

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.2Yes 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i will get the double prime if i will get the derivative 1...?

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

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.0It 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!

kirbykirby
 3 years ago
Best 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 :)

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.2Ok 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

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

tkhunny
 3 years ago
Best 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.

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.2Well 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i 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...

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

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.2No 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 :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ive been memorizing it but i dont know how to use them in solving..

tkhunny
 3 years ago
Best 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok thank you for the answers i will study more...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but for now please help me just last 2 questions..!!

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.0Good call. Show your work, too. This will help us as much as it helps you.

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.2Well 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0my teacher encourages us to use the long method and i just cant get it,,

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.2And 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!!)

tkhunny
 3 years ago
Best 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. :)

terenzreignz
 3 years ago
Best 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?

kirbykirby
 3 years ago
Best 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

tkhunny
 3 years ago
Best 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! :)

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1Chain 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0she 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..!

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

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1Ahh, but sin itself is inside tangent, no?

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.2Yeah 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 [ ]

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

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1leave 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!)

tkhunny
 3 years ago
Best 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Remember 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?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1Well, 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?

kirbykirby
 3 years ago
Best 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

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

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.2That\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

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.2d/dx (x*sinx) = (1)sinx + x(cosx) by the product rule

anonymous
 3 years ago
Best 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? $$u=x\sin x\\y=\tan v\\\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=(\sec^2u)(\sin x+x\cos x)=\sec^2(x\sin x)[\sin x+x \cos x]$$

terenzreignz
 3 years ago
Best 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?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1Also, you don't multiply them, you add them.

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1Hang 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.

anonymous
 3 years ago
Best 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?.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0In 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?

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.2I 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

terenzreignz
 3 years ago
Best 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

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

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1Almost... What's the derivative of sin x?

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

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1Yeah, derivative of sine is cosine Proof on request.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can u give me appropriate solution guys, then ill just study it..

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1Well, 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]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then how will i continue to solve? the answer in the book is 4tanxsec2x

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.2Does your prof require that form of an answer (4tanx sec2x)?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0not exactly it is just an answer in the book.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0another question sir kirbs Y=sec4xtan4x

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.2I 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 :)

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.2You 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:

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.2\[\sec^4x\tan^4x = (\sec^2x)^2\tan^4x\] can you see that?

kirbykirby
 3 years ago
Best 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)

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

kirbykirby
 3 years ago
Best 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

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.2\[=4\tan x \sec^2x\]

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

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.2In 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0=(1+2tan2x+tan4x)−tan4x how did u get this kirb?

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

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.2Ok you know that \[(a+b)^2 = (a+2ab+b^2)\]

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0THANK YOU A LOT KIRB...

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