Need help in solving the differentiation I want to differentiate the following expression with respect to n:
$\frac{d}{dn}[1-{(1-p)^n}^d]$
I do not know which formula to use to differentiate this term with respect to n. Please can anybody help me?
I have tried like this:
$\frac {d}{dn}(1-{(1-p)^n}^d)= -\frac{d}{dn}[{(1-p)^n}^d]=-\frac{d}{dn}[e^{n^dln(1-p)}]=-e^{n^dln(1-p)}.ln(1-p).dn^{d-1}$   Is my answer correct?
Assuming the $d$ in the exponent is a real number and that $p\leq 1$:
$$\frac{d}{dn}[1-(1-p)^{n^d}] = -\frac{d}{dn}(1-p)^{n^d} 
= - \ln(1-p) \times (1-p)^{nd} \times \frac{d}{dn} (n^d) \\ = - n^{d-1}d(1-p)^{nd}\ln(1-p)$$
Which I believe is the same as your answer, although simplified.
$\frac{d}{dn}[1-{(1-p)^n}^d]$ $=-\frac{d}{dn}[{(1-p)^n}^d]$
Let $u={(1-p)^n}^{d}$
therefore $lnu=n^dln(1-p)$
hence  $\frac{d}{dn}[lnu]$=$=\frac{d}{dn}[n^dln(1-p)]$
$\frac{1}{u} \frac{du}{dn}$= $n^d\frac{d}{dn}[ln(1-p)]+ln(1-p)\frac{d}{dn}[n^d]$
$\frac{1}{u} \frac{du}{dn}$= $\frac{n^d}{p-1}\frac{dp}{dn}+ln(1-p)n^{d-1}.d$
$ \frac{du}{dn}$=$u(\frac{n^d}{p-1}\frac{dp}{dn}+ln(1-p)n^{d-1}.d)$
$ \frac{du}{dn}$=${(1-p)^n}^{d}(\frac{n^d}{p-1}\frac{dp}{dn}+ln(1-p)n^{d-1}.d)$
therefore required answer
$- \frac{du}{dn}$ $=-\frac{d}{dn}((1-p)^n)^{d}$=${-(1-p)^n}^{d}(\frac{n^d}{p-1}\frac{dp}{dn}+ln(1-p)n^{d-1}.d)$
If you asssume p to be a constant then $\frac{dp}{dn}=0$
Hence
$- \frac{du}{dn}$ $=-\frac{d}{dn}((1-p)^n)^{d}$=${-(1-p)^n}^{d}.ln(1-p)n^{d-1}.d$