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# Differentiation question watch

1. i did a differentiation question and i wanted to know if its the right answer.

the q is: [(x-1)/(2-x)]^2 and i got [2(x-1)]/[(2-x)^3] sorry dont know how to use latex.

i havent done diff for ages so im a bit rusty and wanted to know if its right and if not, where im going wrong.

thanks much appreciated

btw i used the chain rule and differentiated it as a whole and for the bit inside the brackets i used the quotient rule.
2. I have got the same answer as you, I think you're right.
3. (Original post by Hormonal)
I have got the same answer as you, I think you're right.
thanks for having a look
4. i also have another question and i'd just like someone to check it for me.

its: x=2cos(theta) and y=3sin(theta)

find dy/dx in terms of theta, and the coordinates of the 2 turning points.

Spoiler:
Show
I got dy/dx = 3cos(theta)/-2sin(theta)

then at TP dy/dx = 0 so 3cos(theta) = 0.

so theta must be 90 and/or 270.

I subbed this back into the orig x and y and got the coordinates to be (0,3) and (0,-3).

am I doing this right?

thanks so much
5. (Original post by SurpriseMe)
i also have another question and i'd just like someone to check it for me.

its: x=2cos(theta) and y=3sin(theta)

find dy/dx in terms of theta, and the coordinates of the 2 turning points.

Spoiler:
Show
I got dy/dx = 3cos(theta)/-2sin(theta)

then at TP dy/dx = 0 so 3cos(theta) = 0.

so theta must be 90 and/or 270.

I subbed this back into the orig x and y and got the coordinates to be (0,3) and (0,-3).

am I doing this right?

thanks so much
This is correct, though if I were being picky I'd put the angles in radians
6. (Original post by joostan)
This is correct, though if I were being picky I'd put the angles in radians
lol, to be fair I did do it in radians cos my calc is always in that mode but then for some reason I just felt safer doing it in degrees cos i'm used to the angles at which the cos graph meets the x axis
7. (Original post by SurpriseMe)
lol, to be fair I did do it in radians cos my calc is always in that mode but then for some reason I just felt safer doing it in degrees cos i'm used to the angles at which the cos graph meets the x axis
Meh, it's only a technicality, I know lots of people who do the same thing
8. (Original post by joostan)
Meh, it's only a technicality, I know lots of people who do the same thing
yeah same here

I have another question. How would you diff 2sin^3(e^x)?

would you chnage it to 2[sin(e^x)]^3 and use the chain rule or something?
9. bump.

How would you diff 2sin^3(e^x)?

would you change it to 2[sin(e^x)]^3 and use the chain rule or something?

thanks
10. bump
11. (Original post by SurpriseMe)
bump.

How would you diff 2sin^3(e^x)?

would you change it to 2[sin(e^x)]^3 and use the chain rule or something?

thanks
Product Rule:

y = (u)(v)

dy/dx = (u')(v) + (u)(v')

Okay. Ignore me actually. Wow
12. (Original post by SurpriseMe)
bump.

How would you diff 2sin^3(e^x)?

would you change it to 2[sin(e^x)]^3 and use the chain rule or something?

thanks
Hey, yep it's just the chain rule with three steps:

sin
e^x
[...]^3

Multiply by the differential of all of those. Don't forget to include the 2 at the start as well!

Hope that helps?

Posted from TSR Mobile
13. (Original post by L'Evil Fish)
Product Rule:

y = (u)(v)

dy/dx = (u')(v) + (u)(v')
Why product rule?

(Original post by stuart_aitken)
Hey, yep it's just the chain rule with three steps:

sin
e^x
[...]^3

Multiply by the differential of all of those. Don't forget to include the 2 at the start as well!

Hope that helps?
So it would be 6[sin(e^x)]^2 mutliplied by the differential of what's inside the brackets which is e^xcos(e^x)? Is my differential inside the bracket correct?
14. (Original post by L'Evil Fish)
Product Rule:

y = (u)(v)

dy/dx = (u')(v) + (u)(v')

Okay. Ignore me actually. Wow
What lol? I'm so confused

you're probably way better at maths than i so.... :/
15. (Original post by SurpriseMe)
What lol? I'm so confused

you're probably way better at maths than i so.... :/
I thought we could use it like this:

2Sin^3(e^x)

But then I realised there wasn't an x after Sin^3.

And yes, your differential outside the bracket is right
16. (Original post by L'Evil Fish)
I thought we could use it like this:

2Sin^3(e^x)

But then I realised there wasn't an x after Sin^3.

And yes, your differential outside the bracket is right
Ohhh yeah that would make sense

thank you for checking as well
17. (Original post by SurpriseMe)
Ohhh yeah that would make sense

thank you for checking as well
No problem, my confidence has been knocked a little so I'm wary of answering but I do it anyway
18. (Original post by L'Evil Fish)
No problem, my confidence has been knocked a little so I'm wary of answering but I do it anyway
haha, well as you can see my confidence isn't exactly tip-top otherwise I wouldn't keep asking how to do questions that I'm getting right

well I don't blame you cos if you gave a wrong answer you might have a billion quotes questioning your abilities lol. That's why I stay far away from even answering gcse question
19. (Original post by SurpriseMe)
haha, well as you can see my confidence isn't exactly tip-top otherwise I wouldn't keep asking how to do questions that I'm getting right

well I don't blame you cos if you gave a wrong answer you might have a billion quotes questioning your abilities lol. That's why I stay far away from even answering gcse question
Fair enough

Yeah, lately I've been getting a bit wrong so I haven't answered in a while

What year are you in now?
20. (Original post by L'Evil Fish)
Fair enough

Yeah, lately I've been getting a bit wrong so I haven't answered in a while

What year are you in now?

I'm in year 13, i'm supposed to be an A* student but i think i lack a bit of self-confidence sometimes and i hate being wrong lol :/

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Updated: March 14, 2013
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