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Math Help - Find the differential

  1. #1
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    Find the differential

    I really can not figure out how to proceed, I tried to solve it but apparently the solutions were wrong,

    can someone give me a hint, please...

    Let .
    Find the differential when and dx = 0.3

    thank you
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  2. #2
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    Re: Find the differential

    First, what did you get for the derivative of y with respect to x?
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    Re: Find the differential

    Quote Originally Posted by HallsofIvy View Post
    First, what did you get for the derivative of y with respect to x?
    I tried this way f'(x)= sec^2(4x+7)(4)= (1/cos^2(4x+7))(4)
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    Re: Find the differential

    It always helps if we can actually see what the pictures you post are and if their links aren't considered unsafe by the computer...
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    Re: Find the differential

    what do you mean?
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    Re: Find the differential

    Quote Originally Posted by Prove It View Post
    It always helps if we can actually see what the pictures you post are and if their links aren't considered unsafe by the computer...
    oh, I got it, sorry for that, I thought it would be copied in LaTeX, it's from my univ official webpage, don't worry,

    I still wait some help on it
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    Re: Find the differential

    Is my question so dumb, that's why it's ignored few days in a row?

    Quote Originally Posted by dokrbb View Post
    I tried this way f'(x)= sec^2(4x+7)(4)= (1/cos^2(4x+7))(4)
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    Re: Find the differential

    It might be because your question CAN NOT BE READ.
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    Re: Find the differential

    Quote Originally Posted by Prove It View Post
    It might be because your question CAN NOT BE READ.
    I didn't expect you can`t see them, sorry for that.

    The question is:
    Let y = tan(4x + 7), find the dy when x= 2 and dx=0.3

    I proceded by obtaining the derivative, but I`m completely confused about what should I do further:

    f'(x)= sec^2(4x+7)(4)= (1/cos^2(4x+7))(4)...
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    Re: Find the differential

    Whoever gave you this problem expects you to know the basic definition of "differential"- if y= f(x) then dy= f'(x)dx.
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    Re: Find the differential

    Quote Originally Posted by HallsofIvy View Post
    Whoever gave you this problem expects you to know the basic definition of "differential"- if y= f(x) then dy= f'(x)dx.
    I know the basic deffinition of the differential, I just don't know how to apply it in the case of a trigonometric function

    I tried to solve it, f'(2) = (1/cos^2(4(2)+7))(4) =>

    = (1/cos^2(15))(4) = ... and I recognize I have no idea how to deal with the trig expression cos^2(15)), do I have to consider it as cos(225), which gives me -0,707106, and I just have to plug in the values... (sorry, I haven't done any math since 2002, and I revised a bit of basic trigonometry, but I'm still lost in the trig identities and calculations. I hope though, I would manage to improve that by the time of my final exam...)
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    Re: Find the differential

    That's very basic algebra. No, (cos(x))^2 is NOT cos(x^2). It means exactly what it says- to find (cos(15))^2, find cos(15) (15 radians) and square that.
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    Re: Find the differential

    OK, in this case, I would have

    f'(2) = (1/cos^2(4(2)+7))(4) =>

    = (1/cos^2(15))(4)
    = (1/(.96592582)^2)(4)
    = (1/(.933012)(4)
    = 4.287190;

    after that, f'(2) = 4.287190 *(.3) = 1.286157, that's it?
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    Re: Find the differential

    Quote Originally Posted by dokrbb View Post
    OK, in this case, I would have

    f'(2) = (1/cos^2(4(2)+7))(4) =>

    = (1/cos^2(15))(4)
    = (1/(.96592582)^2)(4)
    = (1/(.933012)(4)
    = 4.287190;

    after that, f'(2) = 4.287190 *(.3) = 1.286157, that's it?
    it still gives me a wrong answer, where is my mistake?
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    Re: Find the differential

    You are taking the cosine of 15 degrees when it should be 15 radians. Put your calculator into radian mode.
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