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    How the heck do you differentiate y=ln(1 + cos2t)? It's on a past paper (Edexcel Jan 03) but I can't find the answers, exam's tomorrow and I'm stuck! Help much appreciated!
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    (Original post by *Joanna*)
    How the heck do you differentiate y=ln(1 + cos2t)? It's on a past paper (Edexcel Jan 03) but I can't find the answers, exam's tomorrow and I'm stuck! Help much appreciated!
    Apply the chain rule, taking u=1+cos2t and then y = ln(u). We have

    dy/du = 1/u = 1/(1+cos2t)
    du/dt = -2sin2t

    Then,

    dy/dt = (dy/du)(du/dt) = -2sin2t/(1+cos2t)
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    (Original post by mikesgt2)
    Apply the chain rule, taking u=1+cos2t and then y = ln(u). We have

    dy/du = 1/u = 1/(1+cos2t)
    du/dt = -2sin2t

    Then,

    dy/dt = (dy/du)(du/dt) = -2sin2t/(1+cos2t)
    Thank you!
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    easy to be honest
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    As you might be able to guess from mike's post, if you have ln (f(x)) this differentiates to f'(x)/f(x), and likewise, e^f(x) differentiates to f'(x) e^f(x). Useful to remember, especially when you have ln of something a bit peculiar.
 
 
 
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