Implicit Differentiation 2
https://isaacphysics.org/questions/implicit_differentiation2?board=d5c15a1b-daa8-429d-882d-53489e220ca5&stage=further_a
[part b]
I tried doing implicit differentiation but I don't know if I just didnt simplify the answer enough or if I made a mistake - could someone please take a look?
Thanks for the help as always.
[part b]
I tried doing implicit differentiation but I don't know if I just didnt simplify the answer enough or if I made a mistake - could someone please take a look?
Thanks for the help as always.
(edited 4 weeks ago)
Working:
Original post by mosaurlodon
Working:
Looks like you have a sign typo when you take the -a over to the right, though I suspect they want you to simply differentiate the original equation, rather than doing the log transformations etc first. Note its not incorrect to write "+ -a/ ..." but youre possibly more likely to make a mistake and writing "- a/... " would be clearer.
Will check your way now (other works) and the answers will be equivalent, just in a different representation.
Edit - looks like isaac doesnt accept your "V-b" solution (even though its equivalent when correct), though its simple enough to transform back to e^(#) at the end if you do it your way.
(edited 4 weeks ago)
thank you for spotting that - so with this new equation:
do I have to take ln() of both sides and then do e^() at the end (or vice versa) or how do I get it in the form dv/dt = e^....
do I have to take ln() of both sides and then do e^() at the end (or vice versa) or how do I get it in the form dv/dt = e^....
Original post by mosaurlodon
thank you for spotting that - so with this new equation:
do I have to take ln() of both sides and then do e^() at the end (or vice versa) or how do I get it in the form dv/dt = e^....
do I have to take ln() of both sides and then do e^() at the end (or vice versa) or how do I get it in the form dv/dt = e^....
Cant say what isaac will/will not accept as its a bit fussy at times, but if you divide top ad bottom by (v-b) and replace RT/(v-b) with the ~exponential in the original equation you should be good.
Note that what youve done is similar to the two ways of deriving the quotient rule for the derivative of
y(x) = f(x)/g(x)
Writing it as f(x)g(x)^(-1) and using the product rule gives the usual form whereas you could say
y(x)g(x) = f(x)
and differentiating gives
y' = (f' - yg')/g
theyre equivallent when you sub for y in the last form, but the initial transformation to simplify the derivative results in an identity transformation in the result.
(edited 4 weeks ago)
Thank you I've got it now 😄
I never knew you could derive quotient rule in 2 ways I've always just done the product rule method.
I never knew you could derive quotient rule in 2 ways I've always just done the product rule method.
Original post by mosaurlodon
Thank you I've got it now 😄
I never knew you could derive quotient rule in 2 ways I've always just done the product rule method.
I never knew you could derive quotient rule in 2 ways I've always just done the product rule method.
Its similar to this question in that you transform the original equation to make it a bit easier to differentiate. As you use the chain rule, youll always get (for this question)
d something /dV * dV/dT = ...
no matter whether you use the original form or the transformed form. As youre interested in dV/dT, you can transform the original equation to make "something" as easy as possible to differentiate.
(edited 4 weeks ago)
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