Use chain rule to find dy/dx for the given value of x. y+ (u-1/u+1)^1/2 u=sqrt x-1; for x=34/9

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Use chain rule to find dy/dx for the given value of x. y+ (u-1/u+1)^1/2 u=sqrt x-1; for x=34/9

Mathematics
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first step I get 1/2 (u-1/u+1)^-1/2 not sure what to do next. I am brain dead at the moment.
what was original function did you replace dx with du du = dx/2sqrt(x-1)
sorry i think im mistaken, ignore what i just said but you will use du

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take what you have and you will multiply it by derivative of (u-1)/(u+1) use quotient rule
what do I do with the ^1/2. I need to bring that infront of the equation and subtract 1 from the 1/2, is that correct?
correct looks like you did that already in the 1st post
ok, after that do I use quotient rule?
yes (f'g -fg')/g^2
ok after that I get (u+1)-(u-1)/(u+1)^2 which ends up being 2/(u+1)^2
correct, now simplify and then you have to multiply by du ans replace u with x
do I need to take derivative of U first?
yes sorry that was what i meant by du (derivative of u) then replace all the u's with sqrt(x-1)
so derivative of u would be 1/2(x-1)^-1/2
correct
do I replace the u with 1/2(x-1)^-1/2 and also replace the x with 34/9 or will that be a different step?
no remember what you assigned u as u=sqrt(x-1) 1/2(x-1)^-1/2 just gets multiplied on the outside Last step will be replacing x with 34/9
so 2/(u+1)^2 * 1/2(x-1)^-1/2 U= sqrt x-1 and x=34/9 ????
yes but dont forget what you had originally 1/2 (u-1/u+1)^-1/2 thats part of it still too
I am lost now would it be 1/2 [(sqrt x-1)-1]/sqrt [(x-1)+1]^-1/2 * 1/2 (34/9-1)^-1/2 ????
2/(u+1)^2 correct just add this part with u = sqrt(x-1) 2/[sqrt(x-1)+1]^2 also replace all the x's with 34/9 i know this one has a lot of parts
this answer really seems wrong, I am sure I did something wrong in my math but I get 2025/136
sorry wrong 2025/128
hmm i worked it out and got 27/320 ill write it out so you can compare \[\frac{dy}{dx} = \frac{1}{(2\sqrt{(34/9)-1})(\sqrt{(34/9)-1}+1)^{2}\sqrt{\frac{\sqrt{(34/9)-1}-1}{\sqrt{(34/9)-1}+1}}}\] 34/9 -1 =25/9 sqrt of that is 5/3
why did you take 1 over everything
because everything had a power of -1/2 right, that means its on the denominator
oh I see now, didn't think of that
so what equation I am using to replace with?
??
it is hard for me to explain to you what I am talking about without showing you my paper or pointing to what you have written. I can see where the (sqrt 34/9)-1-1/(sqrt 34/9)-1+1 came from, the deriviative of original problem 1/2(u-1/u+1)^-1/2. But not the (2sqrt 34/9)-1) (sqrt 34/9 -1+1)^2
ahh it comes from using the chain rule, we are multiplying derivative of original times derivative of inside function ex: f(x) = sqrt(x^2 +1) its a composite function, i say g(x) = x^2+1 f(g) = sqrt(g) f'(g) = 1/2sqrt(g) g'(x) = 2x whats important is the derivative of original is product of these 2 derivatives f'(x) = g'(x)*f'(g) f'(x) = 2x*1/2sqrt(x^2+1) so at the end we replace x^2 +1 back in for g

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