Asked by Tenshi
find y' and y” by implicit differentiation.
2x^3 + 3y^3 = 8
2x^3 + 3y^3 = 8
Answers
Answered by
Steve
6x^2 + 9y^2 y' = 0
y' = -2x^2 / 3y^2
now for y", do it again:
12x + 18y (y')^2 + 9y^2 y" = 0
y" = -2(2x+3y(y')^2) / 3y^2
Now just substitute in y' and you're done
Or, you can use the quotient rule on y':
y' = -2/3 (x^2 / y^2)
y" = -2/3 (2xy^2 - x^2*2yy')/y^4
= -4x/3 (y-xy')/y^3
and now substitute in y'
y' = -2x^2 / 3y^2
now for y", do it again:
12x + 18y (y')^2 + 9y^2 y" = 0
y" = -2(2x+3y(y')^2) / 3y^2
Now just substitute in y' and you're done
Or, you can use the quotient rule on y':
y' = -2/3 (x^2 / y^2)
y" = -2/3 (2xy^2 - x^2*2yy')/y^4
= -4x/3 (y-xy')/y^3
and now substitute in y'
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.