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    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?
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    (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
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    (Original post by elvss567)
    x= 1/y^3 + 2y
    Where did 'ln' go?
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    (Original post by RDKGames)
    Where did 'ln' go?
    I thought differentiating gets rid of ln?
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    (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 x=\ln(y^3+2y) so you want to first find \frac{dx}{dy} by simply differentiating both sides w.r.t y
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    (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 x=\ln(y^3+2y) so you want to first find \frac{dx}{dy} by simply differentiating both sides w.r.t y
    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?
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    (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?
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    (Original post by MadMaths)
    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!!
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    (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 \displaystyle \frac{dy}{dx}=\frac{y^3+2y}{3y^2  +2}. If the mark scheme says \displaystyle \frac{dy}{dx}=\frac{3y^2+2}{y^3+  2y} then it is incorrect.

    You can convince yourself by going to https://www.desmos.com/calculator and typing in your equation x=\ln(y^3+2y). Then let's test for a tangent line, pick the coordinates (\ln 3, 1) on the curve. By our answer, we get the gradient at this point as \frac{3}{5}. By the mark scheme's answer, we get \frac{5}{3}. Clearly one of these is correct, and the other isn't. So type into Desmos y-1=\frac{3}{5}(x-\ln 3) and y-1=\frac{5}{3}(x-\ln 3) and notice how our gradient is the one which gives the tangent line.
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    (Original post by RDKGames)
    Yes you need to reciprocate it to get \displaystyle \frac{dy}{dx}=\frac{y^3+2y}{3y^2  +2}. If the mark scheme says \displaystyle \frac{dy}{dx}=\frac{3y^2+2}{y^3+  2y} then it is incorrect.

    You can convince yourself by going to https://www.desmos.com/calculator and typing in your equation x=\ln(y^3+2y). Then let's test for a tangent line, pick the coordinates (\ln 3, 1) on the curve. By our answer, we get the gradient at this point as \frac{3}{5}. By the mark scheme's answer, we get \frac{5}{3}. Clearly one of these is correct, and the other isn't. So type into Desmos y-1=\frac{3}{5}(x-\ln 3) and y-1=\frac{5}{3}(x-\ln 3) and notice how our gradient is the one which gives the tangent line.
    thankyouu!
 
 
 
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