Sorry the answer choices should be:
-1
11/23
-23/11
-10/3
0
If y5 + 3x2y2 + 5x4 = 49 , then dy/dx at the point (–1, 2) is:
–1
11/23
–23/11
–10/3
0
3 answers
y^5 + 3x^2 y^2 + 5x^4 = 49
5y^4 y' + 6xy^2 + 6x^2y y' + 20x^3 = 0
y' = -(6xy^2+20x^3)/(5y^4+6x^2y)
at (-1,2) y' = -(6(-1)(2^2)+20(-1)^3)/(5(2^4)+6(-1)^2*2) = 11/23
5y^4 y' + 6xy^2 + 6x^2y y' + 20x^3 = 0
y' = -(6xy^2+20x^3)/(5y^4+6x^2y)
at (-1,2) y' = -(6(-1)(2^2)+20(-1)^3)/(5(2^4)+6(-1)^2*2) = 11/23
^^^mans correct: 11/23
0 answers