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Math Help - Implicit Differentiation

  1. #1
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    Implicit Differentiation

    Hello there. I'm new to this forum so I apologise in advance if what I have posted seems a little unclear to you. I just need a little help with the following question.

    If, sinhy = (4sinhx - 3) / (4 + 3sinhx)

    show that,

    dy/dx = -5 / (4 + 3sinhx)

    My final answer contains coshy, and I do not know where to go from there to get the above.

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by Mindless View Post
    Hello there. I'm new to this forum so I apologise in advance if what I have posted seems a little unclear to you. I just need a little help with the following question.

    If, sinhy = (4sinhx - 3) / (4 + 3sinhx)

    show that,

    dy/dx = -5 / (4 + 3sinhx)

    My final answer contains coshy, and I do not know where to go from there to get the above.

    Thanks in advance.
    What were your steps to arrive at cosh(y)?
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  3. #3
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    Your really going to make me type all of that? lol. I basically used the quotient rule which gave me the final result as follows,

    25cosh(x) / cosh(y) (4 + 3sinh(x))^2
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  4. #4
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    Quote Originally Posted by Mindless View Post
    Your really going to make me type all of that? lol. I basically used the quotient rule which gave me the final result as follows,

    25cosh(x) / cosh(y) (4 + 3sinh(x))^2
    It is forum requirement. Plus, it is easier to find a mistake then to type up the whole solution.
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  5. #5
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    That's a very valid point. Okay so these are the steps I took (keeping in mind I used the quotient rule).

    differentiating w.r.t x we get the following,

    cosh(y) dy/dx = [(4 + 3sinh(x))(4cosh(x)) - (4sinh(x) - 3)(3cosh(x)] / (4 + 3sinh(x))^2

    open up the numerator to get,

    cosh(y) dy/dx = 25cosh(x) / (4 + 3sinh(x))^2

    thus,

    dy/dx = 25cosh(x) /cosh(y) (4 + 3sinh(x))^2
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  6. #6
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    Quote Originally Posted by Mindless View Post
    That's a very valid point. Okay so this is the steps I took (keeping in mind I used the quotient rule).

    differentiating w.r.t x we get the following,

    cosh(y) dy/dx = [(4 + 3sinh(x))(4cosh(x)) - (4sinh(x) - 3)(3cosh(x)] / (4 + 3sinh(x))^2

    open up the numerator to get,

    cosh(y) dy/dx = 25cosh(x) / (4 + 3sinh(x))^2

    thus,

    dy/dx = 25cosh(x) /cosh(y) (4 + 3sinh(x))^2
    Have you tried converting the hyperbolics to exponentials and then simplifying?

    \displaystyle\cosh{x}=\frac{e^x+e^{-x}}{2} \ \ \ \ \ \sinh{x}=\frac{e^x-e^{-x}}{2}
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  7. #7
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    No, but the cosh(y) is posing a big problem. I have no idea how to get rid of that, so that I am left with a function which only depends on the variable (x).
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  8. #8
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mindless View Post
    Hello there. I'm new to this forum so I apologise in advance if what I have posted seems a little unclear to you. I just need a little help with the following question.

    If, sinhy = (4sinhx - 3) / (4 + 3sinhx)

    show that,

    dy/dx = -5 / (4 + 3sinhx)

    My final answer contains coshy, and I do not know where to go from there to get the above.

    Thanks in advance.
    Note that \dfrac{\,dy}{\,dx}=\dfrac{25\cosh x}{(\cosh y)(4+3\sinh x)^2}, as you correctly said.

    Now, recall the identity \cosh^2y-\sinh^2y=1. So in your case \cosh y = \sqrt{1+\sinh^2y} = \sqrt{1+\left(\dfrac{4\sinh x-3}{4+3\sinh x}\right)^2}=\ldots

    If you get \cosh y in terms of x, you should be able to get the answer. However, when I did this, I got \dfrac{\,dy}{\,dx}=\dfrac{5}{4+3\sinh x} -- not the negative value you provided.

    I hope this helps!
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  9. #9
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    That was really helpful! yes your right Chris, I also end up with a positive value. Maybe some further fiddling around could bring out the negative five, or maybe the book is incorrect? :P. Anyway, thank you very much for your help. Much appreciated.
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  10. #10
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    Quote Originally Posted by Chris L T521 View Post
     \sqrt{1+\sinh^2y} = \sqrt{1+\left(\dfrac{4\sinh x-3}{4+3\sinh x}\right)^2}=\ldots
    Why is this?
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  11. #11
    Behold, the power of SARDINES!
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    Quote Originally Posted by dwsmith View Post
    Why is this?
    \cosh^2(y)-\sinh^2(y)=1 is an identity for hyperbolic trig functions this gives

    \cosh^2(y)=1+\sinh^2(y) \implies \cosh(y)=\sqrt{1+\sinh^2(y) }

    But in the OP we were given that

    \sinh(y)=\frac{4\sinh(x)-3}{4+3\sinh(x)}
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