Minor typo... there's a missing bracket in the second line, but the third line is correct so that's okay.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.
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.
The derivative of (2/x)^(1/x) is [(1/x)(1- ln(2/x))] * (2/x)^((1/x)-1).
To simplify (2/x)^(1/x), you can rewrite it as e^(ln(2/x) / x), which can be further simplified using the properties of exponents.
The limit of (2/x)^(1/x) as x approaches infinity is 1. This can be shown by taking the natural logarithm of the function and using L'Hopital's rule.
Yes, the second derivative of (2/x)^(1/x) is [(1/x)(1- ln(2/x))] * (2/x)^((1/x)-1) * [(1/x)(1- ln(2/x))] * [(1/x)(1- ln(2/x))-1] - [(1/x^2)(1-ln(2/x))] * (2/x)^((1/x)-1).
The graph of (2/x)^(1/x) is a hyperbola that approaches the x-axis and y-axis as x approaches 0. As x approaches infinity, the graph approaches the line y=1.