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Math Help - a derivative and 1 limit

  1. #1
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    a derivative and 1 limit

    Hello everyone,
    i'm having my exam tomorrow, now doing homework and trying to solve some problems. Here are two that are too hard for me . If anyone could help me solve it...
    Any help will be appreciated

    1 is to find derivative.
    \cos\sqrt{x^2+y^2}=\sin^2x+(\frac{1}{y})
    i've done this:
    \cos\sqrt{x^2+y^2}-\sin^2x-(\frac{1}{y})=0
    then trying to find derivatives of each part

    part 1
    (\cos\sqrt{x^2+y^2})'=-\sin\sqrt{x^2+y^2}*(\sqrt{x^2+y^2})'=
    =-\sin\sqrt{x^2+y^2}*\frac{1}{2\sqrt{x^2+y^2}}*(x^2+  y^2)'=
    =-\frac{\sin\sqrt{x^2+y^2}}{2\sqrt{x^2+y^2}}*(2x*2y)
    part 2 (-\sin^2x)'= can't remember this
    part 3 (\frac{1}{y})'=-\frac{1}{y^2}

    2 limit. \lim_{x\to\infty}(\frac{3x-1}{2x+3})^x not even know how to start
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  2. #2
    MHF Contributor matheagle's Avatar
    Joined
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    Quote Originally Posted by wanqwer View Post
    Hello everyone,
    i'm having my exam tomorrow, now doing homework and trying to solve some problems. Here are two that are too hard for me . If anyone could help me solve it...
    Any help will be appreciated

    1 is to find derivative.
    \cos\sqrt{x^2+y^2}=\sin^2x+(\frac{1}{y})
    i've done this:
    \cos\sqrt{x^2+y^2}-\sin^2x-(\frac{1}{y})=0
    then trying to find derivatives of each part

    part 1 (\cos\sqrt{x^2+y^2})'=-\sin\sqrt{x^2+y^2}*(\sqrt{x^2+y^2})'=
    =-\sin\sqrt{x^2+y^2}*\frac{1}{2\sqrt{x^2+y^2}}*(x^2+  y^2)'=
    =-\frac{\sin\sqrt{x^2+y^2}}{2\sqrt{x^2+y^2}}*(2x*2yy  ')
    part 2 (-\sin^2x)'= can't remember this
    part 3 (\frac{1}{y})'=-\frac{1}{y^2}

    2 limit. \lim_{x\to\infty}(\frac{3x-1}{2x+3})^x not even know how to start

    The first problem is your notation.
    By ' I'm assuming you mean {d\over dx}
    So when you differentiate x wrt s you get one, but when you differentiate
    y wrt x you get {dy\over dx} and not 1.
    NOTE.... I inserted that y' just now in my copy of your solution to 1

    (-\sin^2x)'=-2(\sin x)^1(\cos x)=-\sin (2x)

    (\frac{1}{y})'=-\frac{1}{y^2}{dy\over dx} UNLESS the prime means {d\over  dy}.

    For \lim_{x\to\infty}(\frac{3x-1}{2x+3})^x
    you can divide inside the () and get (1+...).
    BUt it's better to learn how to do this by scratch.
    Let y=(\frac{3x-1}{2x+3})^x, then take log's of both sides, that pulls down the exponent x.
    Work on this and I'll check it later.
    The point to this problem is to make it into a fraction and apply L'Hopitals' Rule.
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