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jayshane

  • 2 years ago

how about differentiate: y=sec²x-tan²x

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  1. terenzreignz
    • 2 years ago
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    Derivative of the sum is the sum of the derivatives.

  2. terenzreignz
    • 2 years ago
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    Although, to be really creative, you can just recall that the derivative of a constant is 0 :D

  3. terenzreignz
    • 2 years ago
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    sec²x = tan²x + 1

  4. wio
    • 2 years ago
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    \[ \sec^2x-\tan^2x = \frac{1}{\cos^2x} - \frac{\sin^2 x}{\cos^2 x} \]

  5. wio
    • 2 years ago
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    So I mean you're gonna end up with y=1

  6. kirbykirby
    • 2 years ago
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    \[\frac{d}{dx}(\sec^2x-\tan^2x)=2\sec x(\sec x \tan x)-2\tan x(\sec^2x) = 2\sec^2x tanx - 2\sec^2x \tan x = 0 \]

  7. kirbykirby
    • 2 years ago
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    = 0 at the end

  8. jayshane
    • 2 years ago
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    the answer in the book is zero...

  9. tkhunny
    • 2 years ago
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    Please listen to @terenzreignz y = 1 Don't miss these sipmle ones.

  10. kirbykirby
    • 2 years ago
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    Sorry it cut off a but, but the last one just cut off "tan x"

  11. terenzreignz
    • 2 years ago
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    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...?

  12. kirbykirby
    • 2 years ago
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    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

  13. jayshane
    • 2 years ago
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    i will get the double prime if i will get the derivative 1...?

  14. terenzreignz
    • 2 years ago
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    It might pay (and save you time) to know by heart your trigonometric identities, too.

  15. tkhunny
    • 2 years ago
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    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!

  16. kirbykirby
    • 2 years ago
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    @jayshane I'm not sure what you mean about getting a double prime? But the derivative of 1 (or any constant) is just 0 :)

  17. kirbykirby
    • 2 years ago
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    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

  18. kirbykirby
    • 2 years ago
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    I usually only think of these identities when I can't "plug and chug" right away lol

  19. tkhunny
    • 2 years ago
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    @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.

  20. kirbykirby
    • 2 years ago
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    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

  21. jayshane
    • 2 years ago
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    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...

  22. terenzreignz
    • 2 years ago
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    Nobody's angry with you @jayshane But, if you have the time, do study trigonometry... :)

  23. kirbykirby
    • 2 years ago
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    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 :)

  24. jayshane
    • 2 years ago
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    ive been memorizing it but i dont know how to use them in solving..

  25. tkhunny
    • 2 years ago
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    @ 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?

  26. jayshane
    • 2 years ago
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    ok thank you for the answers i will study more...

  27. jayshane
    • 2 years ago
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    but for now please help me just last 2 questions..!!

  28. tkhunny
    • 2 years ago
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    Good call. Show your work, too. This will help us as much as it helps you.

  29. kirbykirby
    • 2 years ago
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    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?

  30. jayshane
    • 2 years ago
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    my teacher encourages us to use the long method and i just cant get it,,

  31. kirbykirby
    • 2 years ago
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    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!!)

  32. tkhunny
    • 2 years ago
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    @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. :-)

  33. terenzreignz
    • 2 years ago
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    @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?

  34. kirbykirby
    • 2 years ago
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    @tkhunny Ok good to know we are on good terms :P As for @jayshane, you still need help right

  35. jayshane
    • 2 years ago
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    y=tan(xsinx)

  36. tkhunny
    • 2 years ago
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    @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! :-)

  37. terenzreignz
    • 2 years ago
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    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?

  38. jayshane
    • 2 years ago
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    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..!

  39. terenzreignz
    • 2 years ago
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    Well, @jayshane That's why we're here :) Now, which function is outermost?

  40. jayshane
    • 2 years ago
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    sin?

  41. terenzreignz
    • 2 years ago
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    Ahh, but sin itself is inside tangent, no?

  42. kirbykirby
    • 2 years ago
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    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 [ ]

  43. terenzreignz
    • 2 years ago
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    y=tan(xsinx) And I've already named the outermost function... So you know tan contains all others, what's the derivative of tan?

  44. jayshane
    • 2 years ago
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    sec2x du/dx

  45. terenzreignz
    • 2 years ago
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    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!)

  46. tkhunny
    • 2 years ago
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    \(f(x) = \tan(x\cdot\sin(x))\) -- What a fascinating function. I'm not sure I've ever seen that one in a problem set.

  47. jayshane
    • 2 years ago
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    ok then wats nxt?

  48. oldrin.bataku
    • 2 years ago
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    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?

  49. terenzreignz
    • 2 years ago
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    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?

  50. kirbykirby
    • 2 years ago
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    @oldrin.bataku We have covered the answer for that already :) We are approaching a new question. Sorry to burst your bubble

  51. jayshane
    • 2 years ago
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    -cosx

  52. terenzreignz
    • 2 years ago
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    No... remember xsin x is the product of two functions, now would be a great time to recall your product rule?

  53. kirbykirby
    • 2 years ago
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    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

  54. kirbykirby
    • 2 years ago
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    d/dx (x*sinx) = (1)sinx + x(cosx) by the product rule

  55. oldrin.bataku
    • 2 years ago
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    @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? $$u=x\sin x\\y=\tan v\\\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=(-\sec^2u)(\sin x+x\cos x)=-\sec^2(x\sin x)[\sin x+x \cos x]$$

  56. jayshane
    • 2 years ago
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    (xdsinx)[sinxdx]

  57. terenzreignz
    • 2 years ago
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    @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?

  58. terenzreignz
    • 2 years ago
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    Also, you don't multiply them, you add them.

  59. jayshane
    • 2 years ago
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    sin2x

  60. terenzreignz
    • 2 years ago
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    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.

  61. oldrin.bataku
    • 2 years ago
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    @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?.

  62. jayshane
    • 2 years ago
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    yes im following

  63. oldrin.bataku
    • 2 years ago
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    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?

  64. kirbykirby
    • 2 years ago
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    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

  65. terenzreignz
    • 2 years ago
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    @kirbykirby This is product rule now, though... Eventually we do do away with the u and v, probably

  66. jayshane
    • 2 years ago
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    @oldrin.bataku yeah

  67. kirbykirby
    • 2 years ago
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    Oops I meant the product rule LOL sorry. I am responding to another post and I am getting confused between both posts

  68. jayshane
    • 2 years ago
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    x(-cosx)+sinx(1)

  69. terenzreignz
    • 2 years ago
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    Almost... What's the derivative of sin x?

  70. jayshane
    • 2 years ago
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    -cos x

  71. kirbykirby
    • 2 years ago
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    No not quite. d/dx (sin x) = cos x d/dx (cos x ) = -sin x... yes it's confusing at first

  72. terenzreignz
    • 2 years ago
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    Yeah, derivative of sine is cosine Proof on request.

  73. jayshane
    • 2 years ago
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    can u give me appropriate solution guys, then ill just study it..

  74. terenzreignz
    • 2 years ago
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    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]

  75. jayshane
    • 2 years ago
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    then how will i continue to solve? the answer in the book is 4tanxsec2x

  76. jayshane
    • 2 years ago
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    Y=sec4x-tan4x

  77. kirbykirby
    • 2 years ago
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    Does your prof require that form of an answer (4tanx sec2x)?

  78. jayshane
    • 2 years ago
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    not exactly it is just an answer in the book.

  79. kirbykirby
    • 2 years ago
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    Because I can guarantee you it's the answer :) http://www.wolframalpha.com/input/?i=d%2Fdx+tan%28xsinx%29

  80. jayshane
    • 2 years ago
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    another question sir kirbs Y=sec4x-tan4x

  81. kirbykirby
    • 2 years ago
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    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 :)

  82. kirbykirby
    • 2 years ago
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    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:

  83. kirbykirby
    • 2 years ago
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    \[\sec^4x-\tan^4x = (\sec^2x)^2-\tan^4x\] can you see that?

  84. jayshane
    • 2 years ago
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    yes

  85. kirbykirby
    • 2 years ago
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    \[=(1+\tan^2x)^2-\tan^4x\] \[=(1+2\tan^2x+\tan^4x)-\tan^4x\] by using the identity (a+b)^2 = (a+2ab+b^2)

  86. kirbykirby
    • 2 years ago
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    \[=1+2\tan^2x\]now our problem is much simpler! We find \[\frac{d}{dx}(1+2\tan^2x)=\frac{d}{dx}1+\frac{d}{dx}2\tan^2x\]

  87. kirbykirby
    • 2 years ago
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    \[=0+2\frac{d}{dx}\tan^2x\] \[=2*(2*\tan x)(\sec^2 x) \] 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

  88. kirbykirby
    • 2 years ago
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    \[=4\tan x \sec^2x\]

  89. kirbykirby
    • 2 years ago
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    Think of your box: you have \[[\tan x]^2 = 2[\tan x]*\frac{d}{dx}[\tan x]\]

  90. kirbykirby
    • 2 years ago
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    I hope it's clear

  91. kirbykirby
    • 2 years ago
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    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

  92. jayshane
    • 2 years ago
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    =(1+2tan2x+tan4x)−tan4x how did u get this kirb?

  93. kirbykirby
    • 2 years ago
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    There should be a derivative d/dx in front of [f(x)]^n, sorry my computer was lagging

  94. kirbykirby
    • 2 years ago
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    Ok you know that \[(a+b)^2 = (a+2ab+b^2)\]

  95. jayshane
    • 2 years ago
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    ahh

  96. kirbykirby
    • 2 years ago
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    \[(1+\tan^2x)^2=(1+2(1)(\tan^2x)+(\tan^2x)^2) \].... here we have that \[a = 1, b = \tan^2x\]

  97. jayshane
    • 2 years ago
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    THANK YOU A LOT KIRB...

  98. kirbykirby
    • 2 years ago
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    Your welcome :)

  99. kirbykirby
    • 2 years ago
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    Using the identity \[1+\tan^2x=\sec^2x\]We have \[\sec^2x-\tan^2x=(1+\tan^2x)-\tan^2x\] \[=1\] So, \[\frac{d}{dx}1=0\]

  100. jayshane
    • 2 years ago
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    TY KIRB

  101. kirbykirby
    • 2 years ago
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    :)

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