Finding the Derivative of (2/x)^(1/x)

  • Thread starter Thread starter Nagihiko92
  • Start date Start date
  • Tags Tags
    Differentiate
Nagihiko92
Messages
3
Reaction score
0

Homework Statement


Find (2/x)^(1/x). Simplify your answer.


Homework Equations





The Attempt at a Solution


I let y = (2/x)^(1/x)
lny = ln(2/x)^(1/x)
lny= ln(2/x)/x
However, I get stuck here and don't know whether I should use e on both sides to get y by itself or to use implicit differentiation to get dy/dx directly.
 
Physics news on Phys.org
Nagihiko92 said:
I let y = (2/x)^(1/x)
lny = ln(2/x)^(1/x)
lny= ln(2/x)/x
However, I get stuck here and don't know whether I should use e on both sides to get y by itself or to use implicit differentiation to get dy/dx directly.
Minor typo... there's a missing bracket in the second line, but the third line is correct so that's okay.

If you use e on both sides you just get back what you started with. So that's no good. Try the other. Not sure what you mean by "implicit differentiation". Just differentiate both sides by x.
 
Ok so by using implicit differentiation and simplifying ln(2/x) as ln2 - lnx, I get
(1/y)(dy/dx) = ((x)(0-1/x) - (ln2-lnx)(1))/x
(1/y)(dy/dx) = (-x - ln2 - lnx)/x²
dy/dx = y(-x - ln2 - lnx)/x²

And my final answer turns out
dy/dx = (-xy - yln2 - ylnx)/x²

Did I do something wrong? It looks like I made it even more confusing.
 
Nagihiko92 said:
Ok so by using implicit differentiation and simplifying ln(2/x) as ln2 - lnx, I get
(1/y)(dy/dx) = ((x)(0-1/x) - (ln2-lnx)(1))/x
(1/y)(dy/dx) = (-x - ln2 - lnx)/x²
dy/dx = y(-x - ln2 - lnx)/x²

And my final answer turns out
dy/dx = (-xy - yln2 - ylnx)/x²

Did I do something wrong? It looks like I made it even more confusing.

You can't have y in your answer. That was a variable you introduced. There is also an error in your differentiation of a quotient, and another error later on in the simplifications. But you are on the right track.
 
Ok, so I found my mistakes

(1/y)(dy/dx) = ((x)(-1/x) - (ln(2/x))(1))/x²
(1/y)(dy/dx) = (-1 - ln2 + lnx)/x²
dy/dx = y(-1 - ln2 + lnx)/x²

And if I can't leave y in my answer, I substituted y back in as

dy/dx = ((2/x)^(1/x)(-1 - ln2 + lnx))/x²
Should I factor in the (2/x)^(1/x) or leave it like that?
 
Looks good. Further changes are cosmetic and probably don't matter. Well done.
 
Thread 'Use greedy vertex coloring algorithm to prove the upper bound of χ'
Hi! I am struggling with the exercise I mentioned under "Homework statement". The exercise is about a specific "greedy vertex coloring algorithm". One definition (which matches what my book uses) can be found here: https://people.cs.uchicago.edu/~laci/HANDOUTS/greedycoloring.pdf Here is also a screenshot of the relevant parts of the linked PDF, i.e. the def. of the algorithm: Sadly I don't have much to show as far as a solution attempt goes, as I am stuck on how to proceed. I thought...
Back
Top