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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?

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    $\begingroup$ Can you differentiate $a^x$ with respect to $x$? And if you write it as $e^{x \ln a}$? $\endgroup$
    – Martin R
    Commented May 26, 2021 at 15:11
  • $\begingroup$ Yes. $\frac{d}{dx}(a^x)$ is $ lna.a^x$. $\endgroup$ Commented May 26, 2021 at 15:16
  • $\begingroup$ Correct. Now apply that to your expression. $\endgroup$
    – Martin R
    Commented May 26, 2021 at 15:17
  • $\begingroup$ Ok. I am going to try that now. Thanks. $\endgroup$ Commented May 26, 2021 at 15:18
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    $\begingroup$ $\frac {d}{dn}(1-{(1-p)^n}^d)= -\frac{d}{dn}[{(1-p)^n}^d]=-\frac{d}{dn}[e^{n^dln(1-p)}]=-e^{n^dln(1-p)}.ln(1-p).dn^{d-1}$ Is my answer correct? $\endgroup$ Commented May 26, 2021 at 15:25

2 Answers 2

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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.

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    $\begingroup$ p cannot be equal to 1 it will be strictly less than 1 $\endgroup$ Commented May 26, 2021 at 15:53
  • $\begingroup$ I believe x^t ln(x) goes to 0 for x going to 0 $\endgroup$ Commented May 26, 2021 at 15:58
  • $\begingroup$ x going to zero is a different thing which doesn't necessarily means that x can be zero $\endgroup$ Commented May 26, 2021 at 16:01
  • $\begingroup$ Thank you so much. $\endgroup$ Commented May 26, 2021 at 16:10
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$\frac{d}{dn}[1-{(1-p)^n}^d]$ $=-\frac{d}{dn}[{(1-p)^n}^d]$

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

$- \frac{du}{dn}$ $=-\frac{d}{dn}((1-p)^n)^{d}$=${-(1-p)^n}^{d}.ln(1-p)n^{d-1}.d$

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    $\begingroup$ Thank you so much. $\endgroup$ Commented May 26, 2021 at 16:10
  • $\begingroup$ please upvote if it helped you $\endgroup$ Commented May 26, 2021 at 16:11
  • $\begingroup$ May I know how to do that? $\endgroup$ Commented May 26, 2021 at 16:12
  • $\begingroup$ by clicking on the triangle facing upwards $\endgroup$ Commented May 26, 2021 at 16:13

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