## anonymous 5 years ago (xlny+xy) if i differentiate that with respect to y... i think i get 0

1. anonymous

When you differentiate with respect to y, all you do is treat the $$x$$ variable as a constant. Since $$\ln{y}$$ doesn't differentiate to zero, and neither does $$y$$, you shouldn't get zero.

2. anonymous

ok wat do i get

3. anonymous

cn you show me

4. anonymous

how to do it den

5. anonymous

So the derivative of $$\ln{y}$$ is $$\frac{1}{y}$$, and the derivative of $$y$$ is $$1$$. Does that help get you moving?

6. myininaya

He didn't say partial derivative

7. anonymous

8. myininaya

This could be implicit differiention

9. anonymous

is that rite

10. anonymous

Yep, that's right. Unless myininaya is right about it being implicit differentiation. Are you learning about implicit differentiation now, or partial derivatives?

11. myininaya

If we aren't doing partial differentiation, the answer is x'lny+x/y+x'y+x

12. anonymous

implicit

13. myininaya

Since x is a function of y

14. myininaya

Have you heard of partial derivatives?