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# past paper question watch

1. The diagram shows the curve with equation x=ln(y^3 + 2y).At the point P on the curve, the gradient is 4 and it is given that P is close to the point with coordinates (7.5, 12).

find dy/dx in terms of y. (2 marks)

This is what i done so far:

x= 1/y^3 + 2y
x= (y^3 + 2y)^-1
dx/dy = (y^3 + 2y)^-2 * (3y^2 * 2)

what do i do next?
2. (Original post by elvss567)
The diagram shows the curve with equation x=ln(y^3 + 2y).At the point P on the curve, the gradient is 4 and it is given that P is close to the point with coordinates (7.5, 12).

find dy/dx in terms of y. (2 marks)

This is what i done so far:

x= 1/y^3 + 2y
x= (y^3 + 2y)^-1
dx/dy = (y^3 + 2y)^-2 * (3y^2 * 2)

what do i do next?
finding dy/dx from dx/dy seems like a good next step
3. (Original post by elvss567)
x= 1/y^3 + 2y
Where did 'ln' go?
4. (Original post by RDKGames)
Where did 'ln' go?
I thought differentiating gets rid of ln?
5. (Original post by elvss567)
I thought differentiating gets rid of ln?
Yes but you've not differentiated the LHS at all, and didn't differentiate the RHS correctly.

To start you off, you have so you want to first find by simply differentiating both sides w.r.t
6. (Original post by RDKGames)
Yes but you've not differentiated the LHS at all, and didn't differentiate the RHS correctly.

To start you off, you have so you want to first find by simply differentiating both sides w.r.t
Is this what you mean?

dx/dy = 3y^2 + 2 / y^3 + 2y

surely the equation has to be reciprocated to get dy/dx? however,the mark scheme just states the answer to the question as just that^^ mark scheme obviously is not wrong so where am i going wrong?
7. (Original post by elvss567)
Is this what you mean?

dx/dy = 3y^2 + 2 / y^3 + 2y

surely the equation has to be reciprocated to get dy/dx? however,the mark scheme just states the answer to the question as just that^^ mark scheme obviously is not wrong so where am i going wrong?
Yea the answer is 1 divided by the answer you have. are you sure you written the question right?
Yea the answer is 1 divided by the answer you have. are you sure you written the question right?
yes you're right. I typed up the question wrong!!
9. (Original post by elvss567)
Is this what you mean?

dx/dy = 3y^2 + 2 / y^3 + 2y

surely the equation has to be reciprocated to get dy/dx? however,the mark scheme just states the answer to the question as just that^^ mark scheme obviously is not wrong so where am i going wrong?
Yes you need to reciprocate it to get . If the mark scheme says then it is incorrect.

You can convince yourself by going to https://www.desmos.com/calculator and typing in your equation . Then let's test for a tangent line, pick the coordinates on the curve. By our answer, we get the gradient at this point as . By the mark scheme's answer, we get . Clearly one of these is correct, and the other isn't. So type into Desmos and and notice how our gradient is the one which gives the tangent line.
10. (Original post by RDKGames)
Yes you need to reciprocate it to get . If the mark scheme says then it is incorrect.

You can convince yourself by going to https://www.desmos.com/calculator and typing in your equation . Then let's test for a tangent line, pick the coordinates on the curve. By our answer, we get the gradient at this point as . By the mark scheme's answer, we get . Clearly one of these is correct, and the other isn't. So type into Desmos and and notice how our gradient is the one which gives the tangent line.
thankyouu!

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