C3: Differentiation Question

Watch
#1
0
5 years ago
#2
(Original post by creativebuzz)
...
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.
0
#3
(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
0
5 years ago
#4
(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
0
#5
(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
0
#6
(Original post by aoxa)
I would rearrange it.

1/2 (1/cosy)

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

0
5 years ago
#7
(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.
0
#8
(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)
0
5 years ago
#9
(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)
0
#10
(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)

0
5 years ago
#11
(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.
0
5 years ago
#12
(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.
1
X

new posts
Back
to top
Latest
My Feed

Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Poll

Join the discussion

Yes (65)
21.59%
No (236)
78.41%