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Differentiation question

The eqn of a curve is ln(xy) -y^3 = 1
in the first part i have to find dy/dx which i did, and it is: y/ x(3(y^3) -1)
now second part is find coordinates of point where tangent to curve is parallel to y-axis.
so i know gradient of vertical line is undefined. so to get undefined gradient denominator=0 right? SO i set denominator =0 and get
x (3y^3 -1) = 0 and then i get a value for y. I substitute this value in Original EQN to get value for x. Both answers are correct but I'm not sure whether it's the correct way to do it.
firstly because i also get a value for x=0 when setting denominator =0 but i decided to ignore that x-value.
In a tutorial i saw a similar question, it was solved by setting den=0 and then expressing x or y in terms of the other variable and substituting THAT in original EQN. i need help with this question. How would you proceed with this?
For your problem, once you set the denominator to 0, you find either x = 0 (ruled out because then ln(xy) is undefined) or y^3 = 1. So in this case there's no dependence on x, you can simply read off the value of y and then find x.

But if, for example, you'd found dy/dx = y/(3y^2 - x), then from 3y^2 - x = 0, you wouldn't be *able* to simply deduce a value for x or y - each depends on the other. So in this case you'd *have* to express one in terms of the other, and then substitute back into the original equation.
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papie
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Original post by DFranklin
For your problem, once you set the denominator to 0, you find either x = 0 (ruled out because then ln(xy) is undefined) or y^3 = 1. So in this case there's no dependence on x, you can simply read off the value of y and then find x.

But if, for example, you'd found dy/dx = y/(3y^2 - x), then from 3y^2 - x = 0, you wouldn't be *able* to simply deduce a value for x or y - each depends on the other. So in this case you'd *have* to express one in terms of the other, and then substitute back into the original equation.

thanks for replying, so it's correct if i get a value of y from setting denominator =0 and then substituting that in original eqn to get a value for x?
Original post by papie
thanks for replying, so it's correct if i get a value of y from setting denominator =0 and then substituting that in original eqn to get a value for x?

Yes. Just for 100% clarity, if you, say had a denominator (y - 1)(y-x), you could safely say "y=1" is a solution (and substitute back to find x for that y), but you would also need to find the solutions relating to y = x (again substittuing back to find what x and y mut be) to find all the available solutions.
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papie
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Original post by DFranklin
Yes. Just for 100% clarity, if you, say had a denominator (y - 1)(y-x), you could safely say "y=1" is a solution (and substitute back to find x for that y), but you would also need to find the solutions relating to y = x (again substittuing back to find what x and y mut be) to find all the available solutions.

Thank you so much for clarifying. :smile:

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