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Math Help - quick double check on trig function derivatives

  1. #1
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    quick double check on trig function derivatives

    Ok so I know the derivative of tanx = sec^2(x) so then would the tangent of tan (x^2) = sec^2(x^2)? or sec^4(x). And then the derivative of that would be ?

    I know the derivative of sec(x) = sec(x) tan(x) so sec^2(x) would have derivative sec^2(x)tan^2(x)?

    I'm doing derivative of a derivative. I'm normally not so in need of help with math but I ended up in the hospital for a week and kinda feel behind. Thanks for all the help lately on here.

    Oh, and the original problem was tan^-1(x^2) so then 1/tan(x^2). Just wondering from there so I can double check my answers.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by UMStudent View Post
    Ok so I know the derivative of tanx = sec^2(x) so then would the tangent of tan (x^2) = sec^2(x^2)? or sec^4(x). And then the derivative of that would be ?

    I know the derivative of sec(x) = sec(x) tan(x) so sec^2(x) would have derivative sec^2(x)tan^2(x)?

    I'm doing derivative of a derivative. I'm normally not so in need of help with math but I ended up in the hospital for a week and kinda feel behind. Thanks for all the help lately on here.

    Oh, and the original problem was tan^-1(x^2) so then 1/tan(x^2). Just wondering from there so I can double check my answers.
    Ok, so in the case of something like tan(x^2), we would use the chain rule, so the derivative would be 2x*sec^2(x^2).

    recall that the chain rule says d/dx (f(g(x))) = f '(g(x))*g'(x). what do you think the others will be?
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  3. #3
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    Quote Originally Posted by UMStudent View Post
    Ok so I know the derivative of tanx = sec^2(x) so then would the tangent of tan (x^2) = sec^2(x^2)? or sec^4(x). And then the derivative of that would be ?

    I know the derivative of sec(x) = sec(x) tan(x) so sec^2(x) would have derivative sec^2(x)tan^2(x)?

    I'm doing derivative of a derivative. I'm normally not so in need of help with math but I ended up in the hospital for a week and kinda feel behind. Thanks for all the help lately on here.

    Oh, and the original problem was tan^-1(x^2) so then 1/tan(x^2). Just wondering from there so I can double check my answers.
    Well, let's take tan(x^2) and find the derivative of it to see if you are correct:

    d/dx tan u = sec^2 u du/dx

    Now we plug in x^2 for u

    d/dx tan(x^2) = sec^2(x^2) d(x^2)/dx

    d/dx tan(x^2) = 2x*sec^2(x^2)

    For sec^2(x), recall that it is the same as (sec(x))^2 so that:

    d/dx [u]^2 = 2[u] du/dx

    Now we plug in sec(x) into u

    d/dx sec^2(x) = 2sec(x) d(sec(x))/dx

    d/dx sec^2(x) = 2sec^2(x)tan(x)

    Now, for 1/tan(x^2) we use the quotient rule, just because it's easier for me:

    d/dx 1/tan(x^2) = (1)'(tan(x^2)) - (1)(tan(x^2)' / tan^2(x^2)

    d/dx 1/tan(x^2) = (0)(tan(x^2)) - (1)(2xsec^2(x^2))/tan^2(x^2)

    d/dx 1/tan(x^2) = -2x*sec^2(x^2) / tan^2(x^2)

    And that's your derivative.
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  4. #4
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    Ok so I'm guessing we need to use the product rule on this one. So...

    2x sec^2(x^2)
    2x d/dx sec^2(x^2) + sec^2(x^2) d/dx (2x)
    2x * 2x sec^2(x^2)tan^2(x^2) + sec^2(x^2) *2
    4xsec^2(x^2)tan^2(x^2) + 2 sec^2(x^2)

    Close at all?

    I'm sure I'm wrong because my book gives limited information. If I had some basic rules on derivatives of trig functions it would help like what to do if I have a number x sin ^a(b) or x sin^a(b^c) or just sin^2 x. Do the powers stay whats a constant or do I use other derivative rules the same way I use on say x^3 = 3x^2. There is nothing in the book so I don't know.
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  5. #5
    Super Member Aryth's Avatar
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    Quote Originally Posted by UMStudent View Post
    Ok so I'm guessing we need to use the product rule on this one. So...

    2x sec^2(x^2)
    2x d/dx sec^2(x^2) + sec^2(x^2) d/dx (2x)
    2x * 2x sec^2(x^2)tan^2(x^2) + sec^2(x^2) *2
    4xsec^2(x^2)tan^2(x^2) + 2 sec^2(x^2)

    Close at all?

    I'm sure I'm wrong because my book gives limited information. If I had some basic rules on derivatives of trig functions it would help like what to do if I have a number x sin ^a(b) or x sin^a(b^c) or just sin^2 x. Do the powers stay whats a constant or do I use other derivative rules the same way I use on say x^3 = 3x^2. There is nothing in the book so I don't know.
    Yes, in fact you would use the product rule. So it gives you this:

    d/dx 2xsec^2(x^2) = (d(2x)/dx)(sec^2(x^2)) + (2x)(d(sec^2(x^2)/dx)

    For d/dx (sec(x^2))^2 = 2[u]*d[u]/dx

    For d/dx sec(u) = sec(u)tan(u) du/dx
    d/dx sec(x^2) = 2x*sec(x^2)tan(x^2)

    d/dx (sec(x^2))^2 = 2sec(x^2)*2x sec(x^2)tan(x^2)
    =4x*sec^2(x^2)tan(x^2)

    d/dx 2x*sec^2(x^2) = 2*sec^2(x^2) + 2x(4x*sec^2(x^2)tan(x^2))

    d/dx 2x*sec^2(x^2) = 2*sec^2(x^2) + 8x^2*sec^2(x^2)tan(x^2)

    You can factor out a 2*sec^2(x^2) if you wish to simplify it a bit:

    d/dx 2x*sec^2(x^2) = 2*sec^2(x^2)( 1 + 4x^2*tan(x^2)

    There's your derivative.
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  6. #6
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    Ok so I'm trying to understand this and am starting to get it. So like the derivative of sin^2(x^2) = 2x cos^2(x^2)? Just trying to understand some simple rules like ^2(x^2) is a constant and the derivative of x^2 = 2x so we move that to the front of the statement? Would that be a basic rule I could use so? Or another one the derivative of sin^2 x is cos^2 x so all that changes is the trig function and the power is a constant?

    And if I understand the above correctly then:

    tan^-1(x^2) = 1/tan(x^2) and the derivative of that is
    (using the quotient rule)
    1 * d/dx tan(x^2) - tan(x^2) * d/dx 1 / tan^2(x^2)
    d/dx of 1 = 0 so the left side then pretty much cancles out we are left with
    1 * d/dx tan(x^2) / tan^2(x^2)
    2x sec^2(x^2)/ tan^2(x^2)

    and then i'm lost because i don't know what to do if tan is squared
    d/dx of tan(x) = sec^2(x) so would tan^2(x) = sec^4(x)??

    if correct then I could continue
    2x sec^2(x^2)/2x sec^4(x^2)? and that would be the derivative of 1/tan(x^2)?

    Then I have to take the derivative again. But lets make sure i'm right so far, and I understand some of this basic rules.
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  7. #7
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    Ok I am finally understanding what is going on now I found some other rules in the book I had been previously overlooking so I got the derivative of tan^1(x^2) = -2xsec^2(x^2)/tan^2(x^2) and I understand how I got that but now to take the derivative of that would I have to use the product rule for the top and then also use the quotient rule. Or actually I'm guessing thats way wrong and its just the quotient rule I use again correct? Well actually looking at the problem now I guess I will be using the quotient rule but then using the product rule inside that to find the derivative of like -2x sec^2(x^2) right?
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  8. #8
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    Quote Originally Posted by UMStudent View Post
    Ok I am finally understanding what is going on now I found some other rules in the book I had been previously overlooking so I got the derivative of tan^1(x^2) = -2xsec^2(x^2)/tan^2(x^2) and I understand how I got that but now to take the derivative of that would I have to use the product rule for the top and then also use the quotient rule. Or actually I'm guessing thats way wrong and its just the quotient rule I use again correct? Well actually looking at the problem now I guess I will be using the quotient rule but then using the product rule inside that to find the derivative of like -2x sec^2(x^2) right?
    Yes, to find the derivative of -2x*sec^2(x^2)/tan^2(x^2) you would have to use the quotient rule:

    tan^2(x^2)(d(-2x*sec^2(x^2)/dx) - (2x*sec^2(x^2))*(d(tan^2(x^2)/dx)/(tan^2(x^2))^2

    To get the derivative of -2x*sec^2(x^2) you have to use the product rule:

    (d(-2x)/dx)(sec^2(x^2)) + (-2x)(d(sec^2(x^2))/dx)

    (-2)(sec^2(x^2)) + (-2x)(2sec(x^2)*(2xsec(x^2)tan(x^2))

    -2*sec^2(x^2) -8x*sec^2(x^2)tan(x^2)

    You can factor out a -2*sec^2(x^2) to get:

    d/dx 2x*sec^2(x^2) = -2*sec^2 (1 + 4x*tan(x^2))

    There is your derivative for that, and now we take the derivative of tan^2(x^2):

    d/dx [u]^2 = 2[u] d[u]/dx

    d/dx (tan(x^2))^2 = 2*tan(x^2) (d[tan(x^2)]/dx)

    d/dx tan(x^2) = sec^2(x^2)*2x

    d/dx tan^2(x^2) = 2*tan(x^2)*2x*sec^2(x^2)

    d/dx tan^2(x^2) = 4x*tan(x^2)sec^2(x^2)

    Now we plug them both into the quotient rule:

    tan^2(x^2)(d(-2x*sec^2(x^2)/dx) - (2x*sec^2(x^2))*(d(tan^2(x^2)/dx)/(tan^2(x^2))^2

    tan^2(x^2)(-2*sec^2(x^2) (1 + 4x*tan(x^2)) - (2x*sec^2(x^2))*(4x*tan(x^2)sec^2(x^2))/(tan^2(x^2))^2

    (tan^2(x^2)*-2*sec^2(x^2) + 4x*tan^2(x^2)sec^2(x^2)) - (8x*sec^4(x^2)tan(x^2)/ tan^4(x^2)

    (2x*tan^2(x^2)sec^2(x^2)) - (8x*sec^4(x^2)tan(x^2)) / tan^4(x^2)

    (2x*sec^2(x^2))/(tan^3(x^2)) - (8x*sec^4(x^2))/(tan^3(x^2))

    1/(tan^3(x^2))*(2x*sec^2(x^2)-8x*sec^4(x^2))

    And there's your derivative. If you notice, the more you get the derivative of this stuff, the more complicated it gets.
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