I'm stuck on this problem involving implicit differentiation. The instructions ask me to find y′, the problem is: (x−2y)^3=2y^2−3 So far I've been able to get this far: 3(x−2y)^2(1−2y′)=4y(y′) I've been trying to manipulate it for a while but I can't figure out how to finish the problem properly.

Donna Flynn 2022-07-23 Answered
I'm stuck on this problem involving implicit differentiation.
The instructions ask me to find y , the problem is:
( x 2 y ) 3 = 2 y 2 3
So far I've been able to get this far:
3 ( x 2 y ) 2 ( 1 2 y ) = 4 y ( y )
I've been trying to manipulate it for a while but I can't figure out how to finish the problem properly.
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Answers (1)

Monica Dennis
Answered 2022-07-24 Author has 13 answers
Another way could be to consider the implicit function
f = ( x 2 y ) 3 2 y 2 + 3 = 0
Compute its derivative with respect to x and y
f x = 3 ( x 2 y ) 2
f y = 6 ( x 2 y ) 2 4 y
Now, use the implicit function theorem
d y d x = f x f y = 3 ( x 2 y ) 2 6 ( x 2 y ) 2 4 y = 3 2 × ( x 2 y ) 2 3 ( x 2 y ) 2 + 2 y
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Suppose that a function f ( x ) is differentiable x [ a , b ]. Prove that f ( x ) takes on every value between f ( a ) and f ( b ).

If the above question is a misprint and wants to say "prove that f ( x ) takes on every value between f ( a ) and f ( b )", then I have no problem using the intermediate value theorem here.

If, on the other hand, it is not a misprint, then it seems to me that I can't use the Intermediate value theorem, as I can't see how I am authorised to assume that f ( x ) is continuous on [ a , b ].

Or perhaps there is another way to look at the problem?

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