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 one year ago
When using the Chain Rule, and taking the second piece of derivative (dy/dx of the term for "theta"), would this be correct... (?)
In my particular situation:
The "theta" = (1/y)
So... Because it's not the first step in the chain rule, and I'm "rederiving" as my math teacher says, the constant (1) is affectable. So... Is the following derivation correct?
1/y = (0)/[y^(11)(dy/dx)] = 0/y^(0)(dy/dx) = 0/(1)(dy/dx) = 0/(dy/dx) = (0)
 one year ago
When using the Chain Rule, and taking the second piece of derivative (dy/dx of the term for "theta"), would this be correct... (?) In my particular situation: The "theta" = (1/y) So... Because it's not the first step in the chain rule, and I'm "rederiving" as my math teacher says, the constant (1) is affectable. So... Is the following derivation correct? 1/y = (0)/[y^(11)(dy/dx)] = 0/y^(0)(dy/dx) = 0/(1)(dy/dx) = 0/(dy/dx) = (0)

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amonoconnor
 one year ago
Best ResponseYou've already chosen the best response.1@agent0smith Yes... I know. I'm a thorn in your side ;)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1"theta" = (1/y) ?? I don't understand :(

amonoconnor
 one year ago
Best ResponseYou've already chosen the best response.1Yep, in he function I'm taking the derivative of, the quantity "1/y" is in the "inside" spot, in Chain Rule terminology, or the spot of theta in regard to the trig function also in the function. I have taken the derivative of the whole "outside", finding the derived/ new trig function (theta term untouched), but now I must take the derivative of the theta term itself, and multiply that by the first part; to complete the derivation process, and find the final answer. I'm just wondering if this is in fact the right derivative. :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Ok so for the derivative of the inner function (1/y), You're thinking you get zero for some reason? Did you apply the quotient rule or power rule or something?

amonoconnor
 one year ago
Best ResponseYou've already chosen the best response.1My problem: "ySIN(1/y) = 1  xy"; Find dy/dx. ( Thought this would be helpful to see)

amonoconnor
 one year ago
Best ResponseYou've already chosen the best response.1Umm, no.... I should have, shouldn't I have though? Used the quotient rule?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1You can rewrite it like this:\[\Large \frac{1}{y}\quad=\quad y^{1}\]And apply the power rule as you normally would. I'm not quite sure where your 0 is coming up in the denominator. If you use the quotient rule you should get something like:\[\Large \frac{0y1(1)}{y^2}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1the zero in the numerator, i meant to say :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Oh and a \(\Large y'\) should show up when you chain again.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1But it looks like you understood that part :o i see your dy/dx in there somewhere (although it shouldn't be in the denominator XD heh )

amonoconnor
 one year ago
Best ResponseYou've already chosen the best response.1The "Power Rule"? I'm not familiar with that... This is what we learned in Calc, at least from my teacher: (?) y' = (y^(11))*(dy/dx) = (y^(0))*(dy/dx) = (1)*(dy/dx) = dy/dx

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Is that a y to the first power on the left there?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1You've learned about the chain rule before the power rule...?\[\Large \frac{d}{dx}x^n\quad=\quad n x^{n1}\]That doesn't look familiar? +_+

amonoconnor
 one year ago
Best ResponseYou've already chosen the best response.1Oh... Yes, I know of that. Whoops! I guess we don't know the terminology... I will bring that up to him!

amonoconnor
 one year ago
Best ResponseYou've already chosen the best response.1Ahhhh, I follow your post from earlier... :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Using rules of exponents allows us to write the inner function like this:\[\Large \frac{1}{y}\quad =\quad y^{1}\]From here we don't need to use the quotient rule, we use the power rule instead:\[\Large \frac{d}{dx}y^{1}\quad=\quad 1y^{11}\frac{dy}{dx}\quad=\quad y^{2}\frac{dy}{dx}\]

amonoconnor
 one year ago
Best ResponseYou've already chosen the best response.1Alright.... You're so amazing. I do believe I'm good now. I have what I need, and most importantly, I fully understand what happened, and what I need to do. Thank you so much!

amonoconnor
 one year ago
Best ResponseYou've already chosen the best response.1(Would you mind medaling me?)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1oh got it from there? cool c:

amonoconnor
 one year ago
Best ResponseYou've already chosen the best response.1Yep, I do! Thanks you so much again; you really helped me look at that from a better angle.
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