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    Cosh^-1(x)
    Need to differentiate a function, but don't know how
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    • Community Assistant
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    Look in your formula book?

    Alternatively write \cosh y = \sqrt{5x+3} and take it from there?
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    Easiest way is probobly to take teh cosh of both sides then differentiate implicitly.
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    (Original post by Mr M)
    Look in your formula book?

    Alternatively write \cosh y = \sqrt{5x+3} and take it from there?
    I got this
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    (Original post by megafobia)
    I got this
    Could you show some working?
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    (Original post by Mr M)
    Could you show some working?
    I did it again and I got 1/((5x+2)^(-1/2))

    I have the formula 1/((x^(2)-1)^(-1/2))

    I had (5x+3)^(1/2) that squared (formula) will give 5x+3
    And that would make 1/(5x+3-1)^(-1/2)
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    (Original post by megafobia)
    1/(x^(2)-1)^(-1/2)

    That is the formula of the derivative.

    Then I simply put x into that squared, which was (5x+3)^(-1/2)
    Do you know the Chain Rule?
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    (Original post by Mr M)
    Do you know the Chain Rule?
    Yeah
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    (Original post by megafobia)
    Yeah
    u=\sqrt{5x+3}

    y=\cosh^{-1} u

    \frac{du}{dx}=...

    \frac{dy}{du}=...

    \frac{dy}{dx}=\frac{dy}{du} \times \frac{du}{dx}
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    You can also use this formula

    \displaystyle (f^{-1})^\prime(x) =\frac {1}{f^\prime(f^{-1}(x))}
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    To be honest it is just as quick to implicitly differentiate \cosh y = \sqrt{5x+3}

    \sinh y \frac{dy}{dx}=\frac{5}{2\sqrt{5x  +3}}

    Now \sinh^2 y = \cosh^2 y - 1 ...
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    (Original post by Mr M)
    To be honest it is just as quick to implicitly differentiate \cosh y = \sqrt{5x+3}

    \sinh y \frac{dy}{dx}=\frac{5}{2\sqrt{5x  +3}}

    Now \sinh^2 y = \cosh^2 y - 1 ...
    I tried this and got Name:  ansss.png
Views: 67
Size:  47.6 KB
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    (Original post by megafobia)
    I tried this and got Name:  ansss.png
Views: 67
Size:  47.6 KB
    That isn't right either.

    What's your mathematical background?
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    (Original post by Mr M)
    That isn't right either.

    What's your mathematical background?
    Undergraduate 1st year (with a level maths)

    Name:  hjkhjk.png
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Size:  58.8 KB
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    As you are really struggling here's a worked solution:

    Spoiler:
    Show
    u=\sqrt{5x+3}

    y=\cosh^{-1} u

    \frac{du}{dx}=\frac{5}{2 \sqrt{5x+3}}

    \frac{dy}{du}=\frac{1}{\sqrt{5x+  2}}

    \frac{dy}{dx}=\frac{dy}{du} \times \frac{du}{dx}=\frac{5}{2\sqrt{(5  x+3)(5x+2)}}


    Try to understand each step.
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    (Original post by megafobia)
    Undergraduate 1st year (with a level maths)

    Name:  hjkhjk.png
Views: 62
Size:  58.8 KB
    Undergraduate of what subject?
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    (Original post by Mr M)
    Undergraduate of what subject?
    maths
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    (Original post by Mr M)
    As you are really struggling here's a worked solution:

    Spoiler:
    Show
    u=\sqrt{5x+3}

    y=\cosh^{-1} u

    \frac{du}{dx}=\frac{5}{2 \sqrt{5x+3}}

    \frac{dy}{du}=\frac{1}{\sqrt{5x+  2}}

    \frac{dy}{dx}=\frac{dy}{du} \times \frac{du}{dx}=\frac{5}{2\sqrt{(5  x+3)(5x+2)}}


    Try to understand each step.
    I get everything you've done. Are sure that it is right? I've tried putting it into this uni thing and it says that it is wrong.
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    (Original post by megafobia)
    I get everything you've done. Are sure that it is right? I've tried putting it into this uni thing and it says that it is wrong.
    It look ok to me.

    Perhaps the "uni thing" has just written the same answer in a different form?

    http://www.wolframalpha.com/input/?i...285x%2B3%29%29
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    See if you can manipulate this to get what M has, maple and mathcad
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