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Math Help - Derivative of a Squared Trignometric Function

  1. #1
    Junior Member StarlitxSunshine's Avatar
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    Derivative of a Squared Trignometric Function

    I'm a little confused with this question:

    If y = sec^s (x^2), then  \frac {dy}{dx} =

    1. (A) 2xsec(x^2)tan(x^2)
    2. (B) 4xsec(x^2)tan(x^2)
    3. (C)  sec^2(x^2)tan(x^2)
    4. (D) 4xsec^2(x^2)
    5. (E) 4xsec^2(x^2)tan(x^2)


    See, what I'm not sure about is whether  sec^2(x^2) can be written as  (sec x)^2 or  (sec x^2)^2

    In the first way, I'd get:

    <br />
\frac {d}{dx}[(sec x)^2] = 2secxtanx

    which isn't a choice, but choice (A) is similar...

    And if I use the second way:

    <br />
\frac {d}{dx}[(sec x^2)^2] = 2(sec x^2)(sec x^2tan^2) <br />

    Which isn't in the choices, either. >_<

    Help ? Please & Thank you =)
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  2. #2
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    Quote Originally Posted by StarlitxSunshine View Post
    And if I use the second way:

    <br />
\frac {d}{dx}[(sec x^2)^2] = 2(sec x^2)(sec x^2tan^2) <br />

    Which isn't in the choices, either. >_<
    It is if you allow the chain rule once more and multiply it all by 2x...

    (because when you differentiated the inner function sec(x^2) to get sec(x^2)tan(x^2) you needed the chain rule for that too, because of the 'inner-inner' function x^2.)
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  3. #3
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    Quote Originally Posted by StarlitxSunshine View Post
    I'm a little confused with this question:

    If y = sec^s (x^2), then  \frac {dy}{dx} =

    1. (A) 2xsec(x^2)tan(x^2)
    2. (B) 4xsec(x^2)tan(x^2)
    3. (C)  sec^2(x^2)tan(x^2)
    4. (D) 4xsec^2(x^2)
    5. (E) 4xsec^2(x^2)tan(x^2)


    See, what I'm not sure about is whether  sec^2(x^2) can be written as  (sec x)^2 or  (sec x^2)^2

    In the first way, I'd get:

    <br />
\frac {d}{dx}[(sec x)^2] = 2secxtanx

    which isn't a choice, but choice (A) is similar...

    And if I use the second way:

    <br />
\frac {d}{dx}[(sec x^2)^2] = 2(sec x^2)(sec x^2tan^2) <br />

    Which isn't in the choices, either. >_<

    Help ? Please & Thank you =)
    <br />
sec^2(x^2) = [sec(x^2)][sec(x^2)] = [sec(x^2)]^2

    You'll need to use the chain rule to differentiate this

    2sec(x^2) \cdot 2x \cdot sec(x^2)tan(x^2) = 4xsec^2(x^2)tan(x^2) which corresponds to option E
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by StarlitxSunshine View Post
    I'm a little confused with this question:

    If y = sec^s (x^2), then  \frac {dy}{dx} =

    1. (A) 2xsec(x^2)tan(x^2)
    2. (B) 4xsec(x^2)tan(x^2)
    3. (C)  sec^2(x^2)tan(x^2)
    4. (D) 4xsec^2(x^2)
    5. (E) 4xsec^2(x^2)tan(x^2)


    See, what I'm not sure about is whether  sec^2(x^2) can be written as  (sec x)^2 or  (sec x^2)^2

    In the first way, I'd get:

    <br />
\frac {d}{dx}[(sec x)^2] = 2secxtanx

    which isn't a choice, but choice (A) is similar...

    And if I use the second way:

    <br />
\frac {d}{dx}[(sec x^2)^2] = 2(sec x^2)(sec x^2tan^2) <br />

    Which isn't in the choices, either. >_<

    Help ? Please & Thank you =)
    y = \sec ^2 (x^2)

    y= (\sec (x^2))^2

    \frac{dy}{dx} = 2 \left(\frac{d(\sec (x^2))}{dx} \right)(\sec  (x^2) )^{2-1}

    \frac{dy}{dx} = 2\left(\frac{d(x^2)}{dx}\right)(\sec (x^2)\tan (x^2))(sec(x^2))

    \frac{dy}{dx} = 4x\sec ^2(x^2) \tan x^2
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  5. #5
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    The function y=\sec^2(x^2) if the composition of 3 functions. This means you need to use the chain rule.

    This can be seen as

    y=s^2 \implies \frac{dy}{ds}=2s
    s=sec(t) \implies \frac{ds}{dt}=\sec(t)\tan(t)
    t=x^2 \implies \frac{dt}{dx}=2x

    So the chain rule tells us the derivative is

    \frac{dy}{dx}=\frac{dy}{ds}\frac{ds}{dt}\frac{dt}{  dx}=2s(\sec(t)\tan(t))(2x)

    Now just sub in from above to get

    \frac{dy}{dx}=2\sec(t)\tan(t)(2x)=4x\sec(x^2)\tan(  x^2)
    Last edited by TheEmptySet; November 7th 2009 at 10:40 AM. Reason: edit typo in exponent
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  6. #6
    Junior Member StarlitxSunshine's Avatar
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    Ohh... I understand it now! It's like double chain rule :33

    Thank you very much !

    =)
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  7. #7
    Junior Member StarlitxSunshine's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    The function y=\sec^2(x^2) if the composition of 3 functions. This means you need to use the chain rule.

    This can be seen as

    y=s^2 \implies \frac{dy}{ds}=2s
    s=sec(t) \implies \frac{ds}{dt}=\sec(t)\tan(t)
    t=x^2 \implies \frac{dt}{dx}=2x

    So the chain rule tells us the derivative is

    \frac{dy}{dx}=\frac{dy}{ds}\frac{ds}{dt}\frac{dt}{  dx}=2s(\sec(t)\tan(t))(2x)

    Now just sub in from above to get

    \frac{dy}{dx}=2\sec(t)\tan(t)(2x)=4x^2\sec(x^2)\ta  n(x^2)
    Wait... I didn't see your post when I posted:

    I don't understand the very last step.

    How did you go from:

    \frac{dy}{dx}=2\sec(t)\tan(t)(2x)
    to the equivalent answer ?
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  8. #8
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    Quote Originally Posted by StarlitxSunshine View Post
    Wait... I didn't see your post when I posted:

    I don't understand the very last step.

    How did you go from:

    \frac{dy}{dx}=2\sec(t)\tan(t)(2x)
    to the equivalent answer ?
    He/she defined t= x^2 earlier in their post so it was subbing back x^2 for t
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  9. #9
    Junior Member StarlitxSunshine's Avatar
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    Oh..

    Umm...But that would explain why it was sec(x^2)tan(x^2), but how did he/she get 4x^2 when the remaining was  (2)(2x) = 4x ?
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  10. #10
    Behold, the power of SARDINES!
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    Quote Originally Posted by StarlitxSunshine View Post
    Oh..

    Umm...But that would explain why it was sec(x^2)tan(x^2), but how did he/she get 4x^2 when the remaining was  (2)(2x) = 4x ?

    You are correct there was a typo on my part.

    It has been fixed above.
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  11. #11
    Junior Member StarlitxSunshine's Avatar
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    Ah, ohkay :3

    But now, there are two different answers

    4xsec^2(x^2)tan(x^2) I understand how its done with the product & chain rule because you change the function to (sec x^2)(sec x^2)

    But TheEmptySet has
     4xsec(x^2)tan(x^2) through substituting the parts of the chain rule. So shouldn't the answers be the same, because its essentially the same method ?
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  12. #12
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by StarlitxSunshine View Post
    Ah, ohkay :3

    But now, there are two different answers

    4xsec^2(x^2)tan(x^2) I understand how its done with the product & chain rule because you change the function to (sec x^2)(sec x^2)

    But TheEmptySet has
     4xsec(x^2)tan(x^2) through substituting the parts of the chain rule. So shouldn't the answers be the same, because its essentially the same method ?



    Quote Originally Posted by TheEmptySet View Post
    The function y=\sec^2(x^2) if the composition of 3 functions. This means you need to use the chain rule.

    This can be seen as

    y=s^2 \implies \frac{dy}{ds}=2s
    s=sec(t) \implies \frac{ds}{dt}=\sec(t)\tan(t)
    t=x^2 \implies \frac{dt}{dx}=2x

    So the chain rule tells us the derivative is

    \frac{dy}{dx}=\frac{dy}{ds}\frac{ds}{dt}\frac{dt}{  dx}=2s(\sec(t)\tan(t))(2x)

    Now just sub in from above to get

    \frac{dy}{dx}=2\sec(t)\tan(t)(2x)=4x\sec(x^2)\tan(  x^2)
    he forgot the s the formula is like this

    \frac{dy}{dx}=2s\sec(t)\tan(t)(2x)=2(2x)\sec(x^2)\  sec(x^2)\tan(x^2)[/quote]

    ......
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  13. #13
    Junior Member StarlitxSunshine's Avatar
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    oo0ohkay !

    Thank you so much. After all of that, I should be able to do the question on my own without looking at your working, too !

    Sorry for all the trouble >_<
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