Minimum of a function -approximate value Consider the function below:
$f(x)=\frac{x}{1-\left(1-a^{\frac{1}{x}}\right)^{\frac{1}{d}}}$, $x\geq 1$.
We also have that $0<a<<1$, $d>>1$.
By plotting on Matlab, I can clearly see that this function has an absolute minimum.
However, I am not able to calculate analytically this minimum. I tried via derivation, but the resulting expression has not straightforward zeros.
I don't necessarily need the exact value of the minimum, an approximate value would also be fine.
Anyone can help or suggest a strategy?
 A: For a very tidy approximation, take $\displaystyle x=\frac{1}{\log_{a}\left(\frac{1}{2}\right)}$. To derive this, first observe that $f$ should attain its minimum at the zero of its derivative. But as $f=\frac{u}{v}$ is a quotient, its derivative $f'=\frac{vu'-uv'}{v^2}$ will be zero only if its numerator is. Hence, compute $vu'-uv'$, where $u=x$ and $v=1-\left(1-a^{1/x}\right)^{1/d}$, recalling that since $\left(e^{f(x)}\right)'=f'(x)e^{f(x)}$ and $a^b=e^{b\ln a}$, we must have $\left(a^{1/x}\right)'=-x^{-2}\ln\left(a\right)a^{\frac{1}{x}}$.
We should find
$$vu'-uv' = 1-\left(1-a^{1/x}\right)^{1/d }+\frac{\ln\left(a\right)a^{1/x}}{dx}\left(1-a^{1/x}\right)^{1/d -1}\quad (=0)$$
This seems unappealing but can be swiftly reduced by eliminating the variable $d$ through some slick manipulations. Substitute $h=1/d$ such that for large $d$, we have $h\to0$, and divide through by $h$.
$$-\frac{\left(1-a^{1/x}\right)^{h}-1}{h}+\frac{\ln\left(a\right)a^{1/x}}{x}\left(1-a^{1/x}\right)^{h -1}=0$$
But now recognize the leftmost term as the limit definition of the natural logarithm! $\frac{a^h-1}{h}\to \ln a$. (Equivalently, observe that [treating $x$ as a constant] the fraction on the left is a difference quotient of the function $g(t)=(1-a^{1/x})^t$, explicitly $\frac{g(0+h)-g(0)}{h}\to g'(0)$.) The occurrence of $h$ in the rightmost term can be treated with direct substitution. Hence, in the limit for large $d$, we have
$$-\ln\left(1-a^{1/x}\right)+\frac{\ln\left(a\right)a^{1/x}}{x}\left(1-a^{1/x}\right)^{-1}=0
\\
w\ln\left(w\right)-\left(1-w\right)\ln\left(1-w\right)=0$$
where in the second line, we substitute $w=a^{1/x}$. This last expression gives exactly three zeros, $w=0,\frac12,1$, of which the second is our desired root. Hence, $a^{1/x}=\frac12$ so $x=\frac{1}{\log_{a}\left(\frac{1}{2}\right)}$.
