Question

How do you implicitly differentiate `-1=-y^2x-2xy-ye^(xy)`?

Answer 1

`(dy)/(dx)=(y^2+2y+y^2e^(xy))/(2xy+2x+e^(xy)+xye^(xy))`

Explanation:

When we implicitly differentiate a function `f(y)`, we first differentiate `f(y)` with respect to `y` and then multiply by `(dy)/(dx)`. This comes from chain rule as `(df)/(dx)=(df)/(dy)xx(dy)/(dx)`.

Now using product rule for differentiating `1=-y^2x-2xy-ye^(xy)`, we get

`0=-(2y(dy)/(dx)xx x+y^2)-2(1xxy+x xx(dy)/(dx))-((dy)/(dx)e^(xy)+yxxe^(xy)(1xxy+x xx(dy)/(dx)))` or

`(2xy(dy)/(dx)+y^2)+2(y+x(dy)/(dx))+((dy)/(dx)e^(xy)+ye^(xy)(y+x(dy)/(dx)))=0` or

`(dy)/(dx)(2xy+2x+e^(xy)+xye^(xy))=(y^2+2y+y^2e^(xy))` or

`(dy)/(dx)=(y^2+2y+y^2e^(xy))/(2xy+2x+e^(xy)+xye^(xy))`