Maths question differentiation

When differentiating you forgot to differentiate what was inside the ln (implicit) I would also divide everything by cosx to make the diff easier

Hi, please could I have help with part I, I’m not sure where I’ve gone wrong?
Question: https://ibb.co/vv4NPF5z
Thanks!

Hi, please could I have help with part I, I’m not sure where I’ve gone wrong?
Question: https://ibb.co/vv4NPF5z
Thanks!

Personally, I would first combine the natural logarithms and then split the numerator.
So you’d get y=ln((1+sinx/cosx)= ln(1/cosx + sinx/cosx) = ln(secx + tanx)
And from there’s it’s quite easy to differentiate until you get secx = 1/cosx
Maybe not the best method but easy enough (especially if you’re comfortable with the integration of secx…)
Also, it’s worth noting that your method would’ve been correct but you can’t differentiate ln(cosx/1+sinx) via the chain rule without multiplying by the derivative of the inside (so you’d need the quotient rule on top and then it’s just a lot of nasty algebra…)

When differentiating you forgot to differentiate what was inside the ln (implicit) I would also divide everything by cosx to make the diff easier

Thank you so much it works now, sorry please could you also clarify what you mean when you divide everything by cos x, is it everything in the ln function?

The easiest approach is not to combine the ln terms
dy/dx = cos x/(1 + sin x) + sin x/cos x
then adding gives
dy/dx =(cos^2 x + sin x + sin^2 x)/ (1 + sin x) cos x and we know cos^2 x + sin^2 x = 1
Hence dy/dx = (1 + sin x)/(1 + sin x) cos x = 1/ cos x

The easiest approach is not to combine the ln terms
dy/dx = cos x/(1 + sin x) + sin x/cos x
then adding gives
dy/dx =(cos^2 x + sin x + sin^2 x)/ (1 + sin x) cos x and we know cos^2 x + sin^2 x = 1
Hence dy/dx = (1 + sin x)/(1 + sin x) cos x = 1/ cos x

Sorry, please could you be able to explain how you got the first line?
dy/dx = cos x/(1 + sin x) + sin x/cos x

No, because it's explaining further what others have already done but a smpler approach.
Your method includes more places to go wrong so when you are under time pressure in an exam it wouldn't be an efficient approach.

Fair enough - it’s it a show-that though? Harder to go wrong really either method…

? It's a 'show that' so clear steps needed.
I always encourage students to look for efficient solutions - that's a feature of A* students.

'No, because it's explaining further what others have already done but a smpler approach.
Your method includes more places to go wrong so when you are under time pressure in an exam it wouldn't be an efficient approach.'
The above is what you said before you deleted it - I was saying (especially given that the question was a 'show that') that either method would 'efficiently' and quite clearly lead to the same answer. Takes no more than a couple of lines...
y=ln(1+sinx)-ln(cosx)=ln((1+sinx)/cosx)=ln(secx+sinx/cosx)
Now using standard results (that are given in the formula sheet if not already memorised):
dy/dx=(sec(x)tan(x)+sec^2(x))/(sec(x)+tan(x))=sec(x)((sec(x)+tan(x))/sec(x)+tan(x))=sec(x)=1/cos(x)
Perfectly efficient, 'A* student style' and fast enough in my opinion

Hi, please could I have help with part I, I’m not sure where I’ve gone wrong?
Question: https://ibb.co/vv4NPF5z
Thanks!

where did u get this question from?

'No, because it's explaining further what others have already done but a smpler approach.
Your method includes more places to go wrong so when you are under time pressure in an exam it wouldn't be an efficient approach.'
The above is what you said before you deleted it - I was saying (especially given that the question was a 'show that') that either method would 'efficiently' and quite clearly lead to the same answer. Takes no more than a couple of lines...
y=ln(1+sinx)-ln(cosx)=ln((1+sinx)/cosx)=ln(secx+sinx/cosx)
Now using standard results (that are given in the formula sheet if not already memorised):
dy/dx=(sec(x)tan(x)+sec^2(x))/(sec(x)+tan(x))=sec(x)((sec(x)+tan(x))/sec(x)+tan(x))=sec(x)=1/cos(x)
Perfectly efficient, 'A* student style' and fast enough in my opinion

It’s a booklet of some integration questions my teacher gave, I’ve tried attaching a file but it’s not working. If you would like a copy I can screenshot the questions and send it.

nah thats okay i just wanna know what exam board