## Pinky1234 2 years ago Can anyone help me with this question? Y^2tan(x)=x. Find dy/dx.

1. malcolm11235

you could solve for y then differentiate that way or you could do implicit if you wanted.

2. cwrw238

is this |dw:1359321531751:dw|

3. Pinky1234

yeah we need to do implicit but can you go through with me because this is a new course for me and I am having really tough time with this. I have couple more questions then I ll try by myself.

4. Pinky1234

Yes thats right

5. Pinky1234

looks like no one wants to help me here.

6. malcolm11235

it goes something like |dw:1359321813888:dw|

7. malcolm11235

so then solve for y'

8. malcolm11235

double check what the derivative of tanx is too, i'm not 100% sure

9. Pinky1234

ahan ok and I also have an answer for this that my professor gave which is: \[(1-y ^{2}\sec ^{2}x)/2y \tan (x)\]

10. malcolm11235

that's what you'll get once you rearrage and solve for y'

11. Pinky1234

derivative of tanx is \[\sec ^{2}x\]

12. malcolm11235

okay. now you know how to do it.

13. Pinky1234

no still not sure

14. cwrw238

sorry Pinky - i got called away malcolm is explaining it well

15. Pinky1234

can we go through till the answer

16. Pinky1234

its ok cwrw238

17. malcolm11235

i did go through to the answer. i just didn't rearrange it to solve for dy/dx= y', which is basic algebra, a prerequisite for your calculus course.

18. Pinky1234

hmmmm but can you please also show me to rearrange it if its possible for you?

19. malcolm11235

oops forgot the tanx in there but you just move the one term to the other side then divide by 2ytanx

20. malcolm11235

that last drawing was completely wrong, i don't have time to write it out. it's really basic stuff you should know.

21. malcolm11235

that's not the answr there. thats me saying that y' = dy/dx in case you didn't know that so in the equation, y' = dy/dx

22. Pinky1234

ok

23. malcolm11235

|dw:1359322857339:dw|

24. malcolm11235

that's your full solution. don't come on here and expect people to add 1+1 for you.