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    Would the Dy/DX of Sec squared 3x be 3 Sec sqaured 3x 3Tan Squared 3x
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    (Original post by SunDun111)
    Would the Dy/DX of Sec squared 3x be 3 Sec sqaured 3x 3Tan Squared 3x
    It would be 6tan(3x)sec^2(3x)
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    (Original post by SunDun111)
    Would the Dy/DX of Sec squared 3x be 3 Sec sqaured 3x 3Tan Squared 3x
    No, why do you think that?

    \displaystyle \frac{d}{dx} (\sec 3x)^2 = 2 \times (\sec 3x)^{2-1} \times \frac{d}{dx}(\sec 3x)

    and \frac{d}{dx}(\sec 3x) = 3 \sec 3x \tan 3x
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    (Original post by jjsnyder)
    It would be 2(3secxtanx)^2
    Are you sure?
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    (Original post by Zacken)
    No, why do you think that?

    \displaystyle \frac{d}{dx} (\sec 3x)^2 = 2 \times (\sec 3x)^{2-1} \times \frac{d}{dx}(\sec 3x)

    and \frac{d}{dx}(\sec 3x) = 3 \sec 3x \tan 3x
    Because my teacher didnt teach us the DY/DX of Cot, Cosec and Sec so I am going over it. I thought it would be a 3 Tan 3x but I get it now thank you!
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    (Original post by Zacken)
    Are you sure?
    Aha no totally misread the question :P
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    (Original post by SunDun111)
    Because my teacher didnt teach us the DY/DX of Cot, Cosec and Sec so I am going over it. I thought it would be a 3 Tan 3x but I get it now thank you!
    Just rewrite \sec 3x = \left(\cos 3x\right)^{-1} and then use the chain rule. Same applies for the other two reciprocal functions.
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    (Original post by SunDun111)
    Because my teacher didnt teach us the DY/DX of Cot, Cosec and Sec so I am going over it. I thought it would be a 3 Tan 3x but I get it now thank you!
    they are just reciprocal of tan, sin and cos so just differentiate them in the tan/sin/cos form.

    Edit: whooops
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    (Original post by Zacken)
    Just rewrite \sec 3x = \left(\cos 3x\right)^{-1} and then use the chain rule. Same applies for the other two reciprocal functions.
    Quick question, once I differentiate a trig function, it asks me to show it is an increasing function? What do I do.
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    (Original post by SunDun111)
    Quick question, once I differentiate a trig function, it asks me to show it is an increasing function? What do I do.
    A function f(x) is increasing if its derivative f'(x) > 0 for all x.

    So can you show that your derivative is always positive? Maybe it's a squared quantity or something.
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    (Original post by Zacken)
    A function f(x) is increasing if its derivative f'(x) > 0 for all x.

    So can you show that your derivative is always positive? Maybe it's a squared quantity or something.
    Well its 2Sec2x Tan2x + 2Sec Squared 2x
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    (Original post by SunDun111)
    Well its 2Sec2x Tan2x + 2Sec Squared 2x
    What's the original function?
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    (Original post by Zacken)
    What's the original function?
    The original function was Sec 2x + Tan 2x I had to Differentiate it, it was defined by -pi/4 < x < pi/4
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    (Original post by SunDun111)
    The original function was Sec 2x + Tan 2x I had to Differentiate it, it was defined by -pi/4 < x < pi/4
    Okay, so that got you 2\sec 2x \tan 2x +2\sec^2 2x as your rightly said. Can you show that that is always positive (for the given interval of x, obviously).
 
 
 
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