Question

How do you differentiate `y^2-x^3cos(x) = x^2-y`?

Answer 1

`y'=(2x+3x^2cos(x)-x^3sin(x))/(2y+1)`

Explanation:

Using the chain and the product rule
we get

`2yy'-3x^2c9s(x)+x^3sin(x)=2x-y`
solving for `y'`
`y'(2y+1)=2x+3x^2cos(x)-x^3sin(x)`
so
`y'=(2x+3x^2cos(x)-x^3sin(x))/(2y+1)`
for `2y+1ne 0`