## tmagloire1 one year ago Can anyone help me on this problem regarding implicit differentiation? Thanks http://prntscr.com/7u700y

1. DanJS

when differentiating y, remember to use the chain rule...

2. DanJS

For example, $\frac{ d }{ dx }[y^3] = \frac{ d }{ dy }[y^3]*\frac{ dy }{ dx }$

3. DanJS

Basic chain rule for this $\frac{ d }{ dx } = \frac{ d }{ dy }*\frac{ dy }{ dx }$

4. DanJS

Following the above example, $\frac{ d }{ dx }[y^3] = 3y^2*\frac{ dy }{ dx }$

5. DanJS

Do that for the y terms, and differentiate the x terms normally, then solve for dy/dx

6. tmagloire1

Okay I'll try it and tell you what I get.

7. DanJS

k

8. tmagloire1

Okay I am actually still very confused. Am i supposed to find the second derivative of 5x^2 + y4 = −9 and then use implicit differentiation?

9. DanJS

Yes you want to find the second derivative of the given thing with respect to X, but to take the derivative w.r.t. X of the Y term, you have to use that chain rule... here ill type it out

10. DanJS

$\frac{ d }{ dx }[5x^2]+\frac{ d }{ dx }y^4 = \frac{ d }{ dx }[-9]$ $\frac{ d }{ dx }[5x^2]+\frac{ d }{ dy }y^4*\frac{ dy }{ dx } = \frac{ d }{ dx }[9]$

11. DanJS

now take derivative of y term with respect to y like normal, but you have to put the y' on there too from the chain rule.

12. DanJS

$\frac{ d }{ dx }[5x^2]+\frac{ d }{ dy }y^4*\frac{ dy }{ dx } = \frac{ d }{ dx }[-9]$ $10x^2+4y^3*\frac{ dy }{ dx } = 0$

13. DanJS

Solve that for dy/dx and that is your first derivative...

14. DanJS

Or just remember, if you are taking the derivative of a Y term, tag on a dy/dx next to it ...basically

15. tmagloire1

Okay for the first derivative I got : dy/dx= -10x^2 / 4y^3

16. tmagloire1

And then take the derivative again?

17. DanJS

So you get, $\frac{ dy }{ dx } = \frac{ -10x^2 }{ 4y^3 }$ right

18. DanJS

This time you need the product or the quotient rule and the chain rule...

19. DanJS

I can start typing it out...

20. tmagloire1

Im using the quotient rule

21. DanJS

$\frac{ d }{ dx }\frac{ dy }{ dx } = \frac{ d^2y }{ dx^2 }$ k, ill start typing it

22. tmagloire1

I got : -80xy^3 +120x^2y^2 / (4y^3)^2

23. DanJS

$\frac{ d^2y }{ dx^2 } = \frac{ d }{ dx}\frac{ -5x^2 }{ 2y^3 }$

24. tmagloire1

Oh I think I took the derivative of the wrong thing?

25. tmagloire1

So it would be -20 when you plug in x and y

26. DanJS

$\frac{ d^2y }{ dx^2 }=\frac{ 2y^2*-10x - [-5x*6y^2*\frac{ dy }{ dx } ]}{ 4y^6 }$

27. DanJS

That is the quotient rule , using the chain rule on the y term in the numorator.

28. tmagloire1

So am I supposed to plug in x and y into which equation?

29. DanJS

yes, but notice the second derivative has a term of the first derivative in the numerator... but you know what dy/dx is , so sub that in first

30. DanJS

$\frac{ dy }{ dx } = \frac{ -5x^2 }{ 2y^3 }$ reduced fraction... Sub that in for dyy/dx in the second derivative then plug in the x and y and solve...

31. DanJS

Less writing overall if you want to use y' and y'' instead of dy/dx things

32. DanJS

goodluck, i dont want to calculate it.

33. anonymous

$\large \color{blue}{\checkmark}$

34. tmagloire1

I got -310

35. tmagloire1

Thanks for all the help!

36. DanJS

yw, just get that Leibanitz notation dy/dx crap down, it makes the chain rule and this stuff easier i think .. goodluck

37. tmagloire1

Okay I will try using it more.