Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

how about differentiate: y=sec²x-tan²x

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Derivative of the sum is the sum of the derivatives.
Although, to be really creative, you can just recall that the derivative of a constant is 0 :D
sec²x = tan²x + 1

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question