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marigirl
 one year ago
Implicit differentiation:
I understand that you would need to use it for cases where the question is to differentiate y=x^x
But how do i know when to use it? Could you see the picture i attached below, implicit differentiation was used, however, i could have made it the subject and gone about differentiating it explicitly..
marigirl
 one year ago
Implicit differentiation: I understand that you would need to use it for cases where the question is to differentiate y=x^x But how do i know when to use it? Could you see the picture i attached below, implicit differentiation was used, however, i could have made it the subject and gone about differentiating it explicitly..

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marigirl
 one year ago
Best ResponseYou've already chosen the best response.0When i did this question, i made r the subject by taking the square root of the left side

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3You use it when isolating y is more work than it's worth, sometimes it's even impossible

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3But for your problem, you're saying that end up with a square root which seems different from the correct answer or something?

marigirl
 one year ago
Best ResponseYou've already chosen the best response.0okay, and yes in my question What i did was i did \[r=\sqrt{60hh^2}\] and i thought that was not too bad to differentiate......but the model answers went about implicit differentiating it

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Well you get something that is `not a function` when you solve for r directly,\[\large r=\pm\sqrt{60hh^2}\]Don't forget the +/ when you take the root of a square! :) Maybe that's why they avoided going that way, hmm.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3So doing it implicitly, you end up with:\[\large\rm r'=\frac{30h}{r}\]If we plug r into this equation:\[\large\rm r'=\pm\frac{30h}{\sqrt{30^2(30h)^2}}\]It probably looks more or less the same as what you got, going the other route, yes? :o

marigirl
 one year ago
Best ResponseYou've already chosen the best response.0yes ..so i worried about nothing

marigirl
 one year ago
Best ResponseYou've already chosen the best response.0Thanks :) I appreciate it

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3This MIT video gives an example of this type of thing happening https://youtu.be/5q_3FDOkVRQ?t=9m10s Just in case you wanted to have a better understanding of what is going on, he can explain it much better than I can :D lol

marigirl
 one year ago
Best ResponseYou've already chosen the best response.0I wish to be confident like you in maths one day :) your truly an inspiration

marigirl
 one year ago
Best ResponseYou've already chosen the best response.0\[y=x^{\frac{ m }{ n }}\] how did that become \[y^n=x^m\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Start with a simple example to convince yourself:\[\large\rm y=\sqrt{x}\]If I rewrite this as:\[\large\rm y=x^{1/2}\]In either form, I can see that I need to square to get rid of the root,\[\large\rm y^2=x\]This will happen with any degree of root, your exponent is always the reciprocal of the power on x.\[\large\rm y=\sqrt[3]{x}\qquad\to\qquad y=x^{1/3}\qquad\to\qquad y^3=x\]Err I guess I should be careful the way I say reciprocal, because we really don't care about the numerator on x. So back to the example you gave:\[\large\rm y=x^{m/n}\]To deal with this n'th root, I'm raising both sides to the n'th power,\[\large\rm y^n=\left(x^{m/n}\right)^n\]Rules of exponents gives:\[\large\rm y^n=x^{\frac{m}{n}\cdot n}\]\[\large\rm y^n=x^m\]

marigirl
 one year ago
Best ResponseYou've already chosen the best response.0yes i see it.. thank you so much. im off to watch more MIT videos now :D
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