C3: Differentiation Question Watch

creativebuzz
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#1
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Can someone please guide me through how to answer this question?
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ghostwalker
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(Original post by creativebuzz)
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To get you started - I'm just going out - you should know that dy/dx = 1/(dx/dy), so start by calculating dx/dy.

Edit: Alternatively you can use implicit differentiation.
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creativebuzz
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(Original post by ghostwalker)
To get you started - I'm just going out - you should know that dy/dx = 1/(dx/dy), so start by calculating dx/dy.

Edit: Alternatively you can use implicit differentiation.
Yeah that was the one part I actually knew What I was stuck on was actually differentiating secy/2

And implicit differentiation is C4, not C3
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aoxa
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(Original post by creativebuzz)
Yeah that was the one part I actually knew What I was stuck on was actually differentiating secy/2

And implicit differentiation is C4, not C3
I would rearrange it.

1/2 (1/cosy)

Then differentiate from there
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creativebuzz
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(Original post by aoxa)
I would rearrange it.

1/2 (1/cosy)

Then differentiate from there
Ah I see! I think the point where I got confused was that I assumed that "y/2" was the so-called "angle"! I didn't see it as secy divided by 2
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creativebuzz
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(Original post by aoxa)
I would rearrange it.

1/2 (1/cosy)

Then differentiate from there
Can you spot where I went wrong?

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aoxa
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(Original post by creativebuzz)
Can you spot where I went wrong?


When you differentiate 1/2cosy, you shouldn't put it to the -1 power. Differentiate 2cosy as normal, and just keep the one on the top.

2 cosy doesn't differentiate to -1/2 (cosy)^-2 (siny) though. I have no idea where that came from.
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creativebuzz
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(Original post by aoxa)
When you differentiate 1/2cosy, you shouldn't put it to the -1 power. Differentiate 2cosy as normal, and just keep the one on the top.

2 cosy doesn't differentiate to -1/2 (cosy)^-2 (siny) though. I have no idea where that came from.
But how can you differentiate 1/2cosy? Surely you have to do it to the power of -1 and it's not like it's in the form (u/v)
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aoxa
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(Original post by creativebuzz)
But how can you differentiate 1/2cosy? Surely you have to do it to the power of -1 and it's not like it's in the form (u/v)
You don't do it to the -1 power - it wouldn't work. It might be easier to flip both sides of the fractions, so you get 1/x = 2cosy, then differentiate that.

1/x differentiated would be dy/dx as, as ghost walker said, dy/dx = 1/(dx/dy)
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creativebuzz
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(Original post by aoxa)
You don't do it to the -1 power - it wouldn't work. It might be easier to flip both sides of the fractions, so you get 1/x = 2cosy, then differentiate that.

1/x differentiated would be dy/dx as, as ghost walker said, dy/dx = 1/(dx/dy)
But differentiating 2cosy would give -2siny

If it help, this is what the mark scheme say (I just don't know how they got there, hence I needed help)

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aoxa
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(Original post by creativebuzz)
But differentiating 2cosy would give -2siny

If it help, this is what the mark scheme say (I just don't know how they got there, hence I needed help)

While it's not how I'd attempt the question, what the mark scheme has done, has used the fact that the differential of secy = secytany

It's not how I've ever done a question like that, but that differential is in the formulae book.
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rcmehta
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(Original post by creativebuzz)
But differentiating 2cosy would give -2siny

If it help, this is what the mark scheme say (I just don't know how they got there, hence I needed help)

You did differentiate it correctly - sinx/(cos^2y) is 1/cosx * sinx/cosx which is secx*tanx.

By the way, the y/2 is the angle not (secy)/2. After that they subbed in x=sec(y/2) and used trig formulae at the end to get it into terms of x only. They only made it dy/dx at the final step.

Sorry I can't latex.
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