Find dy/dx at the point (1,1) if,

Question

4x^4 + 3y(x) + 2x^2 y(x)^2 + 3 = 0

Answer 1

If I understand the equation that you wrote down correctly then we need to differentiate in x the expression

4x^4+3y(x)+2(x^2)(y(x)^2)+3=0. We arrive at 16x^3+3y'(x)+4xy(x)^2+4(x^2)y(x)y'(x)=0. Since the point (1,1) belongs to the curve we plug the values x=1 and y=1 into the last expression to obtain that

16+3y'(1)+4+4y'(1)=20+7y'(1)=0. Hence, y'(1)=-20/7.