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    Just a quick question, how would one go about solving this (is there a quicker way than using the exponential form of coshx?)...

    \frac{d}{dx} sin{2x}(cosh{4x})^3
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    There's no need to write cosh(x) in exponential form. Why not just use the product rule with u = \sin 2x and v = (\cosh 4x)^3?
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    There's no need to write cosh(x) in exponential form. Why not just use the product rule with u = \sin 2x and v = (\cosh 4x)^3?
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    Oh dear, I really do have my moments.

    thanks
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    so would the answer be:

     2cos{2x}(cos{4x})^3 + 4sin{2x}sinh{4x}(cosh{4x})^2 ?
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    (Original post by Cavatina)
    so would the answer be:

     2cos{2x}(cos{4x})^3 + 4sin{2x}sinh{4x}(cosh{4x})^2 ?
    It should be \cosh 4x not \cos 4x in the first term (typo?).

    Also, the coefficient of the second term is wrong.
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    (Original post by notnek)
    It should be \cosh 4x not \cos 4x in the first term (typo?).

    Also, the coefficient of the second term is wrong.
    yeah that's a typo, i think it's 12 not 4 in front of the second part, that was just a mistake.
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    (Original post by Cavatina)
    yeah that's a typo, i think it's 12 not 4 in front of the second part, that was just a mistake.
    Correct.
 
 
 
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