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Derivatives proof

  1. Apr 7, 2005 #1
    hi, new to this site.
    i was given this question on a recent test and not quite sure if i got it right.
    it was:

    prove the derivatives of sec^2x is 2sec^3x*sinx.

    i wasn't really sure of the answer. so if anyone could help out, it would be much appreciated.
    thanx :)
  2. jcsd
  3. Apr 7, 2005 #2
    i didn't know to use the quotient rule or the product rule first.
  4. Apr 7, 2005 #3
    Do you know the chain rule? If so, it should be easy.

    If not, I'd probably use the product rule first:

    [tex]\frac{d}{dx}\sec^2{x} = \sec x \frac{d}{dx}\sec x \; + \; \sec x \frac{d}{dx}\sec x = 2 \sec x \frac{d}{dx} \sec x[/tex]

    and from there, you can rewrite it as

    [tex]2 \sec x \frac{d}{dx} \frac{1}{\cos x}[/tex]

    and use the quotient rule.

    There are many other ways to do it, of course, and you could do it by using the quotient rule first if you felt like it.
    Last edited: Apr 7, 2005
  5. Apr 7, 2005 #4
    I get the product/ quotient rule. but with the chian rule doesn't have to be a function within a function. sec^2x isn't a function within a function is it?
  6. Apr 7, 2005 #5

    [tex]f(x) = x^2[/tex]

    [tex]g(x) = \sec x.[/tex]


    [tex] \sec^2 x = f(g(x))[/tex]


    of course, you still have to differentiate [itex]\sec[/itex]~

    You can do that using the chain rule too if you want. Let

    [tex]h(x) = \frac{1}{x}[/tex]


    [tex]u(x) = \cos x.[/tex]


    [tex]\sec x = h(u(x)).[/tex]

    In fact, you can just write

    [tex]\sec^2 x = f(h(u(x)))[/tex]

    and 'peel back the layers' with the chain rule, if you want to.
    Last edited: Apr 7, 2005
  7. Apr 7, 2005 #6


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    Here's the "Chain Rule" method:
    (d/dx){sec^2(x)} = 2*{sec(x)}*(d/dx){sec(x)} =
    = 2*{sec(x)}*(d/dx){cos^(-1)(x)} =
    = 2*{sec(x)}*{(-1)*cos^(-2)(x)}*(d/dx){cos(x)} =
    = 2*{sec(x)}*{(-1)*cos^(-2)(x)}*{(-1)*sin(x)} =
    = 2*{sec(x)}*{(-1)*sec^(2)(x)}*{(-1)*sin(x)} =
    = 2*{sec^3(x)}*{sin(x)}

    Last edited: Apr 7, 2005
  8. Apr 7, 2005 #7
    i sort of get it. but i think i would use the product rule to derive the question.
    thanx for the help i can now sleep at night even though i know i got the question wrong :)
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