Question

Just wondering if someone can check my answer.

Answer 1

using implicit differenation.
\[1+x=\sin(xy^2)\]

for my answer i had:
\[y'=\frac{ 1 }{ cosxy^2(2xy^2) }\]

Answer 2

0 + 1 = cos(xy^2) * (2xy y' + y^2)
2xy cos(xy^2) y' + y^2 cos(xy^2) = 1
2xy cos(xy^2) y' = 1 - y^2 cos(xy^2)
y' = 1 - y^2 cos(xy^2)
---------------
2xy cos(xy^2)

Answer 3

got the same denominator but the product rule forces addtional terms
\[1=\cos(xy^2)*(2xyy' + y^2)\]
\[1= 2xyy'\cos(xy^2) + y^2\cos(xy^2)\]
\[y'=\frac{1-y^2\cos(xy^2)}{2xycos(xy^2)}\]

Answer 4

well thats 2 agreed on this lol!