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# Maths a level differentiation watch

1. The points P(x,y) and Q(x+dx,y+dy) lie on the curve y=3x^2 +2
a) Show that the gradient of the chord PQ is (3(x+dx)^2 - 3x^2)/ dx.
d= delta
so far I have the gradient as y+dy-y/x+dx-x
I then substituted this into y=3x^2+2
y+dy=3(x+dx)^2+2
I'm really unsure of what to do next

---Moved to Maths---
2. (Original post by imaan2121)
The points P(x,y) and Q(x+dx,y+dy) lie on the curve y=3x^2 +2
a) Show that the gradient of the chord PQ is (3(x+dx)^2 - 3x^2)/ dx.
d= delta
so far I have the gradient as y+dy-y/x+dx-x
I then substituted this into y=3x^2+2
y+dy=3(x+dx)^2+2
I'm really unsure of what to do next

---Moved to Maths---
well you know the gradient will be since that is the value of

using P and Qs coordinates and putting them into the curve gives you two equations

1)

2)

can you go from here?

Spoiler:
Show

find dy in terms of x and dx
3. (Original post by DylanJ42)
well you know the gradient will be since that is the value of

using P and Qs coordinates and putting them into the curve gives you two equations

1)

2)

can you go from here?

Spoiler:
Show

find dy in terms of x and dx

How did you get 2)?
4. (Original post by imaan2121)
How did you get 2)?
subbed Qs coordinates into the curve equation since it lies on the curve
5. (Original post by DylanJ42)
subbed Qs coordinates into the curve equation since it lies on the curve
Ah ok thankyou. I understand it up to this point, but still unsure of what to do next
6. (Original post by imaan2121)
Ah ok thankyou. I understand it up to this point, but still unsure of what to do next
well you can find dy in terms of x and dx now, since you have an equation for y, just make a sub

also as you said in your OP, so sub in 1) and 2) in place of y and (y+dy) respectively

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