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# P3 help needed please- evil differentiation! watch

1. 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!
2. (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)
3. (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!
4. easy to be honest
5. 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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