Use logarithmic differentiation to find dy/dx for y=(1+x)^(1/x).
ln y = ln (1+x)^(1/x)
= (1/x) ln (1+x)
(dy/dx) /y = (-1/x^2)(ln(1+x) + (1/x)(1/(1+x)
dy/dx = y [(-1/x^2) ( ln(1+x) + x/(1+x) ]
or (-1/x^2)( ln(1+x) + x/(1+x) ) (1+x)^(1/x)
pretty messy looking ...
To find dy/dx using logarithmic differentiation, we'll take the natural logarithm (ln) of both sides of the equation:
ln(y) = ln((1+x)^(1/x))
Now, we'll use logarithmic properties to simplify the equation. The exponent (1/x) can be moved to the front:
ln(y) = (1/x) * ln(1+x)
Next, we'll differentiate both sides of the equation with respect to x:
(d/dx) ln(y) = (d/dx) [(1/x) * ln(1+x)]
To find the derivative of ln(y), we'll use the chain rule. Let's call it A:
A = ln(y)
Now, differentiate A with respect to x:
(d/dx) A = (d/dy) A * (dy/dx)
Since dA/dx = (d/dx) ln(y), we have:
(d/dx) ln(y) = (dA/dx) = (d/dy) A * (dy/dx)
Substitute back into the equation:
(d/dy) A * (dy/dx) = (1/x) * ln(1+x)
Now, let's find (d/dy) A:
(d/dy) ln(y) = 1/y
So, the equation becomes:
(1/y) * (dy/dx) = (1/x) * ln(1+x)
Now, we can solve for dy/dx:
(dy/dx) = (1/y) * (x/y) * ln(1+x)
Since y = (1+x)^(1/x), we can substitute it back into the equation:
(dy/dx) = (1/(1+x)^(1/x)) * (x/(1+x)^(1/x)) * ln(1+x)
Simplifying further, we have:
(dy/dx) = (x/(1+x)) * ln(1+x) * (1/(1+x)^(1/x))
Therefore, the derivative dy/dx for y = (1+x)^(1/x) is given by the expression:
(dy/dx) = (x/(1+x)) * ln(1+x) * (1/(1+x)^(1/x))