use implicit differentiation.
Remember the product rule and the chain rule:
xy+x^2y^2=x^3y^3
y + xy' + 2xy^2 + 2x^2yy' = 3x^2y^3 + 3x^3y^2y'
y'(x + 2x^2y - 3x^3y^2) = 3x^2y^3-y-2xy^2
y' =
3x^2y^3-y-2xy^2
------------------------
x + 2x^2y - 3x^3y^2
do the other the same way:
xsiny+ycosx=xy
siny + xcosyy' - sinx*y + cosx*y' = y + xy'
...
find dy/dx assuming that y is differentiable with respect to x and x is also differentiable with respect to y.
a)xy+x^2y^2=x^3y^3
b)xsiny+ycosx=xy
4 answers
can you add one more step am not just sure
collect all the y' stuff on one side
y'(x cosy+cosx-x) = y-siny-y sinx
Now just divide to get y' all by itself.
Don't forget your Algebra I now that you're taking calculus!
y'(x cosy+cosx-x) = y-siny-y sinx
Now just divide to get y' all by itself.
Don't forget your Algebra I now that you're taking calculus!
Don't clear me