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# Step 2011 q1 Watch

1. Here's the link to get the paper:
http://www.mathshelper.co.uk/oxb.htm

And here's the link to get the worked solution:
http://www.thestudentroom.co.uk/show....php?t=1697768

I managed to complete Q1 on my own but when I looked at the first step of the worked solution I didn't understand what had been done
Can anyone help explain it please? Thank you
2. (Original post by so it goes)
Can anyone help explain it please? Thank you
Which paper? I, II, or III.
3. (Original post by ghostwalker)
Which paper? I, II, or III.
Ahh, sorry, paper I
4. (Original post by so it goes)
Ahh, sorry, paper I
The first step of the solution is differentiation wrt x
5. (Original post by so it goes)
Here's the link to get the paper:
http://www.mathshelper.co.uk/oxb.htm

And here's the link to get the worked solution:
http://www.thestudentroom.co.uk/show....php?t=1697768

I managed to complete Q1 on my own but when I looked at the first step of the worked solution I didn't understand what had been done
Can anyone help explain it please? Thank you
All they've done is differentiate the equation of the curve w.r.t.x and rearrange, which is the standard way to find the gradient. Are you confused because they've used implicit differentiation to differentiate a function of y and you haven't seen this before?
6. (Original post by davros)
All they've done is differentiate the equation of the curve w.r.t.x and rearrange, which is the standard way to find the gradient. Are you confused because they've used implicit differentiation to differentiate a function of y and you haven't seen this before?
Yes, that's it! Thanks, I just watched a quick tutorial on implicit differentiation and had a go at the question, it makes sense now thank you very much
7. (Original post by joostan)
The first step of the solution is differentiation wrt x
Thank you
8. (Original post by so it goes)
Yes, that's it! Thanks, I just watched a quick tutorial on implicit differentiation and had a go at the question, it makes sense now thank you very much
No problem!

I haven't attempted this particular question in detail, but there are often several ways of attacking questions about tangents to quadratic curves - e.g. you can use differentiation, or if you're given the equation for the tangent substitute it into the equation for the curve and use a condition for equal roots

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