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Math Help - differentation

  1. #1
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    differentation

    differentate :ln(sinx cos x) answer provided2 cot 2x)

    dy/du = 1/u du/dx= -sinx cosx

    -sin x cos x/( ln [(sinx cosx)] )

    -sinx cosx / (1/sinx + 1/cos x )

    stucked
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  2. #2
    Super Member Matt Westwood's Avatar
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    Best way to start is by using the rule of logs:

    y = \ln \left({\sin x \cos x}\right) = \ln \left({\sin x}\right) + \ln \left({\cos x}\right)

    Diffing that is easy, leads to

    \frac {dy}{dx} = \frac {\cos x}{\sin x} - \frac {\sin x} {\cos x}

    I left it as sin & cos to make the next stage easier, i.e. put it over a common denominator.

    Then you have an expression on the top and the bottom for which the double angle formulae for sin 2x and cos 2x can be used, and you're done.
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  3. #3
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    did you mean

    cos^2 x - sin^2 x/ cos x sin x

    so i stucked at here.

    btw cos x sin x mean wht? cos^2 x - sin^2 x = cos 2x
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  4. #4
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by qweiop90 View Post
    did you mean

    cos^2 x - sin^2 x/ cos x sin x

    so i stucked at here.

    btw cos x sin x mean wht? cos^2 x - sin^2 x = cos 2x
    So you got the formula for \cos 2x - now what's the formula for \sin 2x then?
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  5. #5
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    2 sinx cos x
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  6. #6
    Super Member Matt Westwood's Avatar
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    so where's the problem?
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  7. #7
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    2 sin x cos x is not equivalent to sin x cos x. lets say sin x cos x = sin x

    cos 2x / sinx , izzit possible to be 2cot2x?
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  8. #8
    Super Member Matt Westwood's Avatar
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    uh?!?!?

    Do you know how to multiply and divide by 2?

    \sin 2x = 2 \sin x \cos x = 2 \cos x \sin x

    Theeeerreefoooooooore...

    \cos x \sin x = \frac {\sin 2x}{2}
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  9. #9
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    hehehe ok . i get it. thx
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