anonymous
  • anonymous
how about differentiate: y=sec²x-tan²x
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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terenzreignz
  • terenzreignz
Derivative of the sum is the sum of the derivatives.
terenzreignz
  • terenzreignz
Although, to be really creative, you can just recall that the derivative of a constant is 0 :D
terenzreignz
  • terenzreignz
sec²x = tan²x + 1

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anonymous
  • anonymous
\[ \sec^2x-\tan^2x = \frac{1}{\cos^2x} - \frac{\sin^2 x}{\cos^2 x} \]
anonymous
  • anonymous
So I mean you're gonna end up with y=1
kirbykirby
  • kirbykirby
\[\frac{d}{dx}(\sec^2x-\tan^2x)=2\sec x(\sec x \tan x)-2\tan x(\sec^2x) = 2\sec^2x tanx - 2\sec^2x \tan x = 0 \]
kirbykirby
  • kirbykirby
= 0 at the end
anonymous
  • anonymous
the answer in the book is zero...
tkhunny
  • tkhunny
Please listen to @terenzreignz y = 1 Don't miss these sipmle ones.
kirbykirby
  • kirbykirby
Sorry it cut off a but, but the last one just cut off "tan x"
terenzreignz
  • terenzreignz
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...?
kirbykirby
  • kirbykirby
Yes there are many ways to approach this one. You can re-write 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
  • anonymous
i will get the double prime if i will get the derivative 1...?
terenzreignz
  • terenzreignz
It might pay (and save you time) to know by heart your trigonometric identities, too.
tkhunny
  • tkhunny
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!
kirbykirby
  • kirbykirby
@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
  • kirbykirby
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
kirbykirby
  • kirbykirby
I usually only think of these identities when I can't "plug and chug" right away lol
tkhunny
  • tkhunny
@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
  • kirbykirby
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
anonymous
  • anonymous
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...
terenzreignz
  • terenzreignz
Nobody's angry with you @jayshane But, if you have the time, do study trigonometry... :)
kirbykirby
  • kirbykirby
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 :)
anonymous
  • anonymous
ive been memorizing it but i dont know how to use them in solving..
tkhunny
  • tkhunny
@ 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
  • anonymous
ok thank you for the answers i will study more...
anonymous
  • anonymous
but for now please help me just last 2 questions..!!
tkhunny
  • tkhunny
Good call. Show your work, too. This will help us as much as it helps you.
kirbykirby
  • kirbykirby
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?
anonymous
  • anonymous
my teacher encourages us to use the long method and i just cant get it,,
kirbykirby
  • kirbykirby
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!!)
tkhunny
  • tkhunny
@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
  • terenzreignz
@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
  • kirbykirby
@tkhunny Ok good to know we are on good terms :P As for @jayshane, you still need help right
anonymous
  • anonymous
y=tan(xsinx)
tkhunny
  • tkhunny
@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
  • terenzreignz
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?
anonymous
  • anonymous
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..!
terenzreignz
  • terenzreignz
Well, @jayshane That's why we're here :) Now, which function is outermost?
anonymous
  • anonymous
sin?
terenzreignz
  • terenzreignz
Ahh, but sin itself is inside tangent, no?
kirbykirby
  • kirbykirby
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 [ ]
terenzreignz
  • terenzreignz
y=tan(xsinx) And I've already named the outermost function... So you know tan contains all others, what's the derivative of tan?
anonymous
  • anonymous
sec2x du/dx
terenzreignz
  • terenzreignz
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!)
tkhunny
  • tkhunny
\(f(x) = \tan(x\cdot\sin(x))\) -- What a fascinating function. I'm not sure I've ever seen that one in a problem set.
anonymous
  • anonymous
ok then wats nxt?
anonymous
  • anonymous
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?
terenzreignz
  • terenzreignz
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?
kirbykirby
  • kirbykirby
@oldrin.bataku We have covered the answer for that already :) We are approaching a new question. Sorry to burst your bubble
anonymous
  • anonymous
-cosx
terenzreignz
  • terenzreignz
No... remember xsin x is the product of two functions, now would be a great time to recall your product rule?
kirbykirby
  • kirbykirby
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
kirbykirby
  • kirbykirby
d/dx (x*sinx) = (1)sinx + x(cosx) by the product rule
anonymous
  • anonymous
@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? $$u=x\sin x\\y=\tan v\\\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=(-\sec^2u)(\sin x+x\cos x)=-\sec^2(x\sin x)[\sin x+x \cos x]$$
anonymous
  • anonymous
(xdsinx)[sinxdx]
terenzreignz
  • terenzreignz
@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
  • terenzreignz
Also, you don't multiply them, you add them.
anonymous
  • anonymous
sin2x
terenzreignz
  • terenzreignz
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.
anonymous
  • anonymous
@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
  • anonymous
yes im following
anonymous
  • anonymous
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?
kirbykirby
  • kirbykirby
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 u-notation once you master it that way
terenzreignz
  • terenzreignz
@kirbykirby This is product rule now, though... Eventually we do do away with the u and v, probably
anonymous
  • anonymous
@oldrin.bataku yeah
kirbykirby
  • kirbykirby
Oops I meant the product rule LOL sorry. I am responding to another post and I am getting confused between both posts
anonymous
  • anonymous
x(-cosx)+sinx(1)
terenzreignz
  • terenzreignz
Almost... What's the derivative of sin x?
anonymous
  • anonymous
-cos x
kirbykirby
  • kirbykirby
No not quite. d/dx (sin x) = cos x d/dx (cos x ) = -sin x... yes it's confusing at first
terenzreignz
  • terenzreignz
Yeah, derivative of sine is cosine Proof on request.
anonymous
  • anonymous
can u give me appropriate solution guys, then ill just study it..
terenzreignz
  • terenzreignz
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]
anonymous
  • anonymous
then how will i continue to solve? the answer in the book is 4tanxsec2x
anonymous
  • anonymous
Y=sec4x-tan4x
kirbykirby
  • kirbykirby
Does your prof require that form of an answer (4tanx sec2x)?
anonymous
  • anonymous
not exactly it is just an answer in the book.
kirbykirby
  • kirbykirby
Because I can guarantee you it's the answer :) http://www.wolframalpha.com/input/?i=d%2Fdx+tan%28xsinx%29
anonymous
  • anonymous
another question sir kirbs Y=sec4x-tan4x
kirbykirby
  • kirbykirby
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 :)
kirbykirby
  • kirbykirby
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:
kirbykirby
  • kirbykirby
\[\sec^4x-\tan^4x = (\sec^2x)^2-\tan^4x\] can you see that?
anonymous
  • anonymous
yes
kirbykirby
  • kirbykirby
\[=(1+\tan^2x)^2-\tan^4x\] \[=(1+2\tan^2x+\tan^4x)-\tan^4x\] by using the identity (a+b)^2 = (a+2ab+b^2)
kirbykirby
  • kirbykirby
\[=1+2\tan^2x\]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
  • kirbykirby
\[=0+2\frac{d}{dx}\tan^2x\] \[=2*(2*\tan x)(\sec^2 x) \] 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
  • kirbykirby
\[=4\tan x \sec^2x\]
kirbykirby
  • kirbykirby
Think of your box: you have \[[\tan x]^2 = 2[\tan x]*\frac{d}{dx}[\tan x]\]
kirbykirby
  • kirbykirby
I hope it's clear
kirbykirby
  • kirbykirby
In general for a "power" you have like \[[f(x)]^n=n[f(x)]^{n-1}*\frac{d}{dx}f(x)\] I hope I am not confusing you by writing this
anonymous
  • anonymous
=(1+2tan2x+tan4x)−tan4x how did u get this kirb?
kirbykirby
  • kirbykirby
There should be a derivative d/dx in front of [f(x)]^n, sorry my computer was lagging
kirbykirby
  • kirbykirby
Ok you know that \[(a+b)^2 = (a+2ab+b^2)\]
anonymous
  • anonymous
ahh
kirbykirby
  • kirbykirby
\[(1+\tan^2x)^2=(1+2(1)(\tan^2x)+(\tan^2x)^2) \].... here we have that \[a = 1, b = \tan^2x\]
anonymous
  • anonymous
THANK YOU A LOT KIRB...
kirbykirby
  • kirbykirby
Your welcome :)
kirbykirby
  • kirbykirby
Using the identity \[1+\tan^2x=\sec^2x\]We have \[\sec^2x-\tan^2x=(1+\tan^2x)-\tan^2x\] \[=1\] So, \[\frac{d}{dx}1=0\]
anonymous
  • anonymous
TY KIRB
kirbykirby
  • kirbykirby
:)

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