Question

How do you implicitly differentiate `2=e^(xy)-cosy +xy^3`?

Answer 1

`(dy)/dx=-(ye^(xy)+y^3)/(xe^(xy)+siny+3xy^2)`

Explanation:

`(d(2))/dx=(d(e^(xy)-cosy+xy^3))/dx`

`0=(d(e^(xy)))/dx-(d(cosy))/dx+(d(xy^3))/dx`

`0=(d(xy))/dx*e^(xy)-((dy)/dx)(-siny)+((dx)/dx*y^3)+x(d(y^3))/dx`

`0=(y+x*(dy)/dx)*e^(xy)+((dy)/dx*siny)+y^3+3xy^2*(dy)/dx`

`0=ye^(xy)+xe^(xy)(dy)/dx+(dy)/dx*siny+y^3+3xy^2*(dy)/dx`

Collecting all similar monomials including `(dy)/dx`:

`0=xe^(xy)*(dy)/dx+(dy)/dx*siny+3xy^2*(dy)/dx+ye^(xy)+y^3`

`0=(dy)/dx*(xe^(xy)+siny+3xy^2)+(ye^(xy)+y^3)`

`-(dy)/dx*(xe^(xy)+siny+3xy^2)=ye^(xy)+y^3`

`(dy)/dx=-(ye^(xy)+y^3)/(xe^(xy)+siny+3xy^2)`