## anonymous 5 years ago find dx/dy for y^2 = 5x+2

1. anonymous

Have you heard of implicit differentiation?

2. anonymous

yes i was wondering what the answer was as my teacher provided no answers. just wanted to double check =]

3. anonymous

Well, anyway, you need to take the derivative with respect to x on both sides. The left-hand side, being a function of x (i.e. y = y(x) will be differentiated using the chain rule). $y^2=5x+2 \rightarrow \frac{d}{dx}y^2=\frac{d}{dx}(5x+2) \rightarrow2yy'=5$$y'=\frac{5}{2y} \rightarrow y'=\frac{5}{2\sqrt{5x+2}}, -\frac{5}{2\sqrt{5x+2}}$

4. anonymous

5. anonymous

what is the difference between dx/dy and dy/dx?

6. anonymous

dy/dx is the rate of change in y with respect to x, and dx/dy is the rate of change of x with respect to y. In the first instance, we're measuring the change in y as a function of x (i.e. *we* change x and the function moves because of it). In the second, we're measuring the change in x as a function of y (i.e. *we* change y and x is forced to move because of it).

7. anonymous

Does that make sense?

8. anonymous

yes that does thank you very much! so for the question you just helped me out with you had found dx/dy right ?

9. anonymous

or did you find dy/dx?

10. anonymous

No...what you're technically doing is called an 'operation' in mathematics. You can move from one statement to another in mathematics and perform operations at each step, the only catch is that if you perform an operation on one side of an equation, you need to perform the operation on the other side.

11. anonymous

So, we needed the derivative of this equation with respect to x. I said we needed to take the derivative with respect to x on both sides.

12. anonymous

When we did it to the right-hand side, it was easy, since$\frac{d}{dx}(5x+2)=5$...you're used to that.

13. anonymous

But when we did it to the left-hand side, it wasn't so obvious as to what we were doing. What we're doing in implicit differentiation is using the chain rule. So, on the left-hand side,

14. anonymous

$\frac{d}{dx}y^2=\frac{dy^2}{dy}.\frac{dy}{dx}=2y \frac{dy}{dx}=2yy'$

15. anonymous

Do you see how we've used the chain rule? Nothing dodgy happened.

16. anonymous

awwww i see. hahahas i'm really stupid at maths. Well THANK YOU SO SO MUCH for your time and patience in explaining! =D

17. anonymous

np ;) and I'm pretty sure you're not stupid!

18. anonymous

You should be able to do your other problem now.

19. anonymous

THANK YOU THANK YOU !