## mathcalculus 2 years ago can someone explain to me Implicit Differentiation: if y^3 = 25x^2, determine dx/dt when x = 5 and dy/dt = 1

1. mathcalculus

these are the steps but i don't understand them :/ 3y^2*dy/dt = 50x*dx/dt When x = 5, y^3 = 25*5^2 y^3 = 625 y = cube root 625 = 8.55 3(8.55)^3*1 = 50*5*dx/dt 1875 = 250dx/dt 7.5 = dx/dt

2. mathcalculus

isn't y=2.92402?

3. mathcalculus

i did this on the calculator: sqrt (625)^(1/2)

4. mathcalculus

(1/3)**

5. mathcalculus

@satellite73

6. agent0smith

Let's start with this part... $\Large \frac{ d }{ dt } y^3$use the chain rule to differentiate it... so bring down the exponent, reduce the exponent by one, then multiply by the derivative of y w. respect to t $\Large 3 y^2 *\frac{ dy }{ dt }$

7. mathcalculus

so I can't replace the dy/dt by 1 since it's given?

8. mathcalculus

@agent0smith

9. agent0smith

Well yes but i was seeing if you understood the differentiation... don't worry about plugging in numbers till later.

10. mathcalculus

and 3y^2 is already the derivative, why do we need to use the chain rule?

11. mathcalculus

@agent0smith

12. agent0smith

Because you have to multiply by the derivative of y with respect to t.

13. agent0smith

3y^2 is NOT the derivative of y^3, UNLESS you're just differentiating with respect to y.

14. tpmys

normally we have y dependent on x here y, and x both are dependent on some t, so we dont know what the function is so instead of y^3 >>> 3y^2 we have the chain rule y^3 >>> 3y^2 * y'

15. agent0smith

It's the same with dy/dx... the derivative of y with respect to x is not just 1... it's 1*dy/dx

16. mathcalculus

ok... :/ it's a little confusing..

17. ranga

agent0smith explained it nicely: $\frac{ d }{ dy }(y ^{3}) =3y ^{2}$ But $\frac{ d }{ dt }y ^{3} = \frac{ d }{ dy }y ^{3}\frac{ dy }{ dt } = 3y ^{2}\frac{ dy }{ dt }$

18. agent0smith

It's the same thing with the chain rule $\huge \frac{ d }{ dx } f(x)^n = n*f(x)^{n-1} * f \prime (x)$ you have to always multiply by the derivative at the end.

19. mathcalculus

why can't we plug in the numbers right away if it's given?

20. agent0smith

You can... you're probably just better off not doing it until you really know what you're doing.

21. mathcalculus

right i understand that... @ranga

22. mathcalculus

well i was taught to plug in.. that's why.

23. mathcalculus

that's the confusing part.

24. agent0smith

Fair enough... but you can't plug in until you have this step: 3y^2*dy/dt = 50x*dx/dt

25. mathcalculus

exactly what i have right now.

26. mathcalculus

so dy/dt is 1...

27. mathcalculus

in the steps above, why is this: 3(8.55)^3*1 = 50*5*dx/dt?

28. mathcalculus

y=8.55? y is suppose to be 2.92402.

29. mathcalculus

and it's raised to the 3rd, and why not 2?

30. agent0smith

When x = 5, y^3 = 25*5^2 y^3 = 625 y = cube root 625 = 8.55 i think the next line after that is a mistake and it should by to the power of 2

31. mathcalculus

so y=2.92402 is correct?

32. agent0smith

y^3 = 625 y = cube root 625 = 8.55

33. mathcalculus

how?

34. mathcalculus

i did this in the calculator: sort(625)^(1/3)

35. mathcalculus

sqrt*

36. agent0smith

And you'll get 8.55... 2.94 cubed is close to 3^3 which is 27. Not 625.

37. mathcalculus

i swear im getting 2.92402

38. mathcalculus

i plugged it in exactly like iwrote it.

39. agent0smith

sort(625)^(1/3) why are you taking the square root of a cubed root...

40. agent0smith

y^3 = 625 y = 625^(1/3)

41. mathcalculus

OHHH

42. mathcalculus

i thought the sqrt could do cube root.

43. mathcalculus

if we did the exponent also.

44. agent0smith

That would mean ((625^(1/3))^(1/2) = 625^(1/6)

45. mathcalculus

thank you! i got it !