# Find the inflection point on the equation $\frac{1}{x^2y^3}=7$

Find the inflection point of the equation $\frac{1}{x^2y^3}=7$

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What have you done? – jspecter Dec 17 '11 at 0:31
@jspecter - i don't understand the use of partial derivative and i don't have some sample with solution(i don't have money to buy books), so i try to crete some exemple – Victor Dec 17 '11 at 0:34
Victor, people don't really care about your personal problems and this is not the place to share them or hope for pity. jspecter asked you to show what did you try to do to solve this problem, don't reply about how you cannot afford books. – Asaf Karagila Dec 17 '11 at 7:51

First notice that $x^2y^3=1/7$. Then it follows that $2xy^3+3x^2y^2(dy/dx)=0$ so that $dy/dx=-2y/3x$. Then, $d^2y/dx^2=((-6x(dy/dx))+6y)/(9x^2)$. Then find points where the second derivative is equal to zero or does not exist.