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'
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