Question

How do you implicitly differentiate `2=e^ysinx-x^2y^3`?

Answer 1

I found: `(dy)/(dx)=(2xy^3-e^ycos(x))/(e^ysin(x)-3x^(2)y^2)`

Explanation:

You need to consider `y` as a function of `x` and so differentiate it accordingly. For example:
if you have `y^2` differentiating you'll get `2y(dy)/(dx)` to consider the fact that `y` represents a function of something!
in our case:
`0=e^y(dy)/(dx)sin(x)+e^ycos(x)-2xy^3-x^(2)3y^2(dy)/(dx)`
Here I used the Product Rule and every time I differentiated `y` I needed to include `(dy)/(dx)`:

collect `(dy)/(dx)`:
`(dy)/(dx)[e^ysin(x)-x^(2)3y^2]=2xy^3-e^ycos(x)`
and:
`(dy)/(dx)=(2xy^3-e^ycos(x))/(e^ysin(x)-x^(2)3y^2)`