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# Differentiation watch

1. (iii) Find the equation of the tangent to the curve at the point where x = -1.

when x = -1

(-1,-1)

What do I do after this??

y + 4x + 3 = 0
2. (Original post by ckfeister)
(iii) Find the equation of the tangent to the curve at the point where x = -1.

when x = -1

(-1,-1)

What do I do after this??

y + 4x + 3 = 0
Can you please post the full question?
3. (Original post by ckfeister)
...

Also as noted above, post the the full question.
4. (Original post by notnek)
Can you please post the full question?

3.(i) Find in terms of p the gradient of the line PQ, where P is the point (p, p4) on the curve y = x4, and Q is the point on the curve where x = p + h.

(ii) Use your answer to (i) to find the gradient of the tangent to the curve at P.

(iii) Find the equation of the tangent to the curve at the point where x = -1.

----
3i is
3ii is
5. (-1)^4 = 1 and not -1

Then sub X = -1 and Y = 1 and M into the equation

Y - Y1 = M (X - X1)

Y - 1 = -4 (X + 1)

Y - 1 = -4X - 4

Y + 4X + 3 = 0

Posted from TSR Mobile
7. If the tangent at x=p has gradient g (i.e. here g = p^3), then it must have equation

y = gx + c for some constant c.

You choose c so that gx+c passes through the point on the curve where x = p.
8. (Original post by RDKGames)

Also as noted above, post the the full question.
I did below.

(Original post by TheMightyAugur)

Then sub X = -1 and Y = 1 and M into the equation

Y - Y1 = M (X - X1)

Y - 1 = -4 (X + 1)

Y - 1 = -4X - 4

Y + 4X + 3 = 0

Posted from TSR Mobile
So x = -1 also means p = -1??
9. (Original post by ckfeister)
So x = -1 also means p = -1??
Yes. That would be the point P if you refer to the context of the question. You have derived the gradient at this point from first principles.
10. (Original post by RDKGames)
Yes. That would be the point P if you refer to the context of the question. You have derived the gradient at this point from first principles.
(Original post by DFranklin)
If the tangent at x=p has gradient g (i.e. here g = p^3), then it must have equation

y = gx + c for some constant c.

You choose c so that gx+c passes through the point on the curve where x = p.
(Original post by notnek)
Can you please post the full question?
(Original post by TheMightyAugur)

Then sub X = -1 and Y = 1 and M into the equation

Y - Y1 = M (X - X1)

Y - 1 = -4 (X + 1)

Y - 1 = -4X - 4

Y + 4X + 3 = 0

Posted from TSR Mobile

4. A and B lie on the curve y = 1/x at the points where x = -2 and x = -2 + h, respectively.

(i) Find the gradient of the line AB.

How do I finish this off?

Never seen questions like this before...
11. (Original post by ckfeister)
4. A and B lie on the curve y = 1/x at the points where x = -2 and x = -2 + h, respectively.

(i) Find the gradient of the line AB.

A(-2, -1/2)

How do I finish this off?

Never seen questions like this before...
12. Rather than go straight for "what is m", you may want to calculate separately

= the change in x (so -2 - (-2-h)) here.

= the change in y (so ) here.

Get both of these down to single fraction expressions and then combine to find .

It's not really any different from what you're doing but the individual expressions are simpler and you're a lot less likely to get confused than if you have expressions like (it's easier to LaTeX, too )
13. (Original post by RDKGames)
I've got to

m = -h/2 + h/-2+h

What do I do from here??
(i) 1/ 2(h - 2)
14. (Original post by ckfeister)
I've got to

m = -h/2 + h/-2+h
15. (Original post by ckfeister)
I've got to

m = -h/2 + h/-2+h

What do I do from here??
That's not right... where did the h come from in the numerator?
16. (Original post by RDKGames)
That's not right... where did the h come from in the numerator?

I've never seen any questions or examples on these, idk what I'm doing but I did ..

17. (Original post by ckfeister)

I've never seen any questions or examples on these, idk what I'm doing but I did ..

If the first line you were dividing by h, and it has mysteriously changed into you multiplying by h. This is wrong.
18. (Original post by DFranklin)
If the first line you were dividing by h, and it has mysteriously changed into you multiplying by h. This is wrong.
a/b/c = ac/b correct? as a/b / 1/c = a/b * c/1 = a/b *c/1 = ac/b
19. (Original post by ckfeister)

I've never seen any questions or examples on these, idk what I'm doing but I did ..

Yeah those two are not equal to one another. If you multiply the numerator of the main fraction by h then you must multiply the denominator of the main fraction by h, thus giving you on the main denominator which not what you want. Also it should be on your main denominator by the looks of it anyway.

To carry on correctly, simplify first.
20. (Original post by ckfeister)
a/b/c = ac/b correct? as a/b / 1/c = a/b * c/1 = a/b *c/1 = ac/b
What yo have is (a/b)/c. This is the same as a/(bc), not ac/b.

Note that division is not associative, so once you have more than one division you should use brackets to avoid ambiguity. (And so I'm not even going to try to follow your attempted justification which has more divisions than the current Labour party...)

Edit: Note also that there's a reason I advised you to find (and simplify) and separately, as you would then largely avoid constructs like this that you evidently find confusing.

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