Question

x^2y^2=cos(x^2+y^2)
solve for y-prime
:)

Answer 1

x^2y^2 = cos(x^2+y^2)


d/dx(x^2y^2) = d/dx(cos(x^2+y^2))


2xy^2 + 2x^2y*y' = -sin(x^2+y^2)*(2x+2y*y')


2xy^2 + 2x^2y*y' = -2x*sin(x^2+y^2)-2y*y'*sin(x^2+y^2)


2x^2y*y'+2y*y'*sin(x^2+y^2) = -2x*sin(x^2+y^2)-2xy^2


y'(2x^2y+2y*sin(x^2+y^2)) = -2x*sin(x^2+y^2)-2xy^2


y' = (-2x*sin(x^2+y^2)-2xy^2)/(2x^2y+2y*sin(x^2+y^2))

Answer 2

@jim_thompson5910
what happened here:-2x*sin(x^2+y^2)-2y*y'*sin(x^2+y^2) the 3rd line
thats where i got stocked

Answer 3

On the left side, i use the product rule. On the right side, I used the chain rule.

Answer 4

oh and wherever you're differentiating y wrt x, then you're using the chain rule as well

Answer 5

oh ok! thanks alot :)

Answer 6

you're welcome