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Is it possible to find $n^{th}$ derivative of $frac{1}{e^{ax}+b}$?


$n^{th}$ derivative of $cot x$What is the proof of integral theorem i.e area under curve is given by anti derivative?Struggling to find the second derivative of this function's first derivativelargest interval over which the general solution is definedA fence of $y$ ft is $x$ ft from a wall, find shortest ladder using trigonometryDouble derivative w.r.t x and y neededQuick question regarding $int frac{cos x + sin x}{sin 2x},dx$Find the derivative of: $(f^{-1})'(0)$ with $f(x)=int_{1}^{x}cos(cos(t))dt$derivative of integralFind the $lim_{n to 0} (cos(x)+tan^2(x))^{csc(x)}$Differentiating $x = sec^2{3y}$ for $d^2y/dx^2$













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$begingroup$



Is it possible to find $n^{th}$ derivative of $frac{1}{e^{ax}+b}$ ?




I am introduced to $n^{th}$ differentiation and leibniz theorem as a first year undergraduate.
But what i observe is there are some explicit formula for the $n^{th}$ derivative of $sin x$,$cos x$,$log(ax+b)$,$frac{1}{ax+b}$, etc but why there isn't any method given for $sec x$, $csc x$ etc.




If the highlighted question can be answered then it is possible to find the $n^{th}$ derivative of secx.
How ?
$$y=sec x=frac{1}{cos x}=frac{2 e^{ix}}{e^{2ix}+1}$$
Now one can use lebniz theorem of $n^{th}$ differentiation to find the n derivative of product of two functions whose general derivative is known. Now here $e^{ix}$ whose derivative can be found easily. The latter part $frac{1}{e^{2ix+1}}$ is not known.






calculus algebra-precalculus derivatives







share|cite|improve this question











$endgroup$












  • $begingroup$
    If $y=frac{1}{e^{ax}+b}$, then $y'= a b y^2 - a y$. Perhaps this helps.
    $endgroup$
    – lhf
    Mar 12 at 0:16










  • $begingroup$
    How good is that , differentiation of $n$ times of this will yield its n-th derivative in terms of n-1th or less.
    $endgroup$
    – M Desmond
    Mar 12 at 15:20












  • $begingroup$
    For $cot x$, see math.stackexchange.com/questions/1701540/…
    $endgroup$
    – lhf
    Mar 13 at 21:21


















2












$begingroup$



Is it possible to find $n^{th}$ derivative of $frac{1}{e^{ax}+b}$ ?




I am introduced to $n^{th}$ differentiation and leibniz theorem as a first year undergraduate.
But what i observe is there are some explicit formula for the $n^{th}$ derivative of $sin x$,$cos x$,$log(ax+b)$,$frac{1}{ax+b}$, etc but why there isn't any method given for $sec x$, $csc x$ etc.




If the highlighted question can be answered then it is possible to find the $n^{th}$ derivative of secx.
How ?
$$y=sec x=frac{1}{cos x}=frac{2 e^{ix}}{e^{2ix}+1}$$
Now one can use lebniz theorem of $n^{th}$ differentiation to find the n derivative of product of two functions whose general derivative is known. Now here $e^{ix}$ whose derivative can be found easily. The latter part $frac{1}{e^{2ix+1}}$ is not known.






calculus algebra-precalculus derivatives







share|cite|improve this question











$endgroup$












  • $begingroup$
    If $y=frac{1}{e^{ax}+b}$, then $y'= a b y^2 - a y$. Perhaps this helps.
    $endgroup$
    – lhf
    Mar 12 at 0:16










  • $begingroup$
    How good is that , differentiation of $n$ times of this will yield its n-th derivative in terms of n-1th or less.
    $endgroup$
    – M Desmond
    Mar 12 at 15:20












  • $begingroup$
    For $cot x$, see math.stackexchange.com/questions/1701540/…
    $endgroup$
    – lhf
    Mar 13 at 21:21
















2












2








2


0



$begingroup$



Is it possible to find $n^{th}$ derivative of $frac{1}{e^{ax}+b}$ ?




I am introduced to $n^{th}$ differentiation and leibniz theorem as a first year undergraduate.
But what i observe is there are some explicit formula for the $n^{th}$ derivative of $sin x$,$cos x$,$log(ax+b)$,$frac{1}{ax+b}$, etc but why there isn't any method given for $sec x$, $csc x$ etc.




If the highlighted question can be answered then it is possible to find the $n^{th}$ derivative of secx.
How ?
$$y=sec x=frac{1}{cos x}=frac{2 e^{ix}}{e^{2ix}+1}$$
Now one can use lebniz theorem of $n^{th}$ differentiation to find the n derivative of product of two functions whose general derivative is known. Now here $e^{ix}$ whose derivative can be found easily. The latter part $frac{1}{e^{2ix+1}}$ is not known.






calculus algebra-precalculus derivatives







share|cite|improve this question











$endgroup$





Is it possible to find $n^{th}$ derivative of $frac{1}{e^{ax}+b}$ ?




I am introduced to $n^{th}$ differentiation and leibniz theorem as a first year undergraduate.
But what i observe is there are some explicit formula for the $n^{th}$ derivative of $sin x$,$cos x$,$log(ax+b)$,$frac{1}{ax+b}$, etc but why there isn't any method given for $sec x$, $csc x$ etc.




If the highlighted question can be answered then it is possible to find the $n^{th}$ derivative of secx.
How ?
$$y=sec x=frac{1}{cos x}=frac{2 e^{ix}}{e^{2ix}+1}$$
Now one can use lebniz theorem of $n^{th}$ differentiation to find the n derivative of product of two functions whose general derivative is known. Now here $e^{ix}$ whose derivative can be found easily. The latter part $frac{1}{e^{2ix+1}}$ is not known.






calculus algebra-precalculus derivatives




calculus algebra-precalculus derivatives






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 11 at 19:26







M Desmond

















asked Mar 11 at 19:17









M DesmondM Desmond

1637




1637












  • $begingroup$
    If $y=frac{1}{e^{ax}+b}$, then $y'= a b y^2 - a y$. Perhaps this helps.
    $endgroup$
    – lhf
    Mar 12 at 0:16










  • $begingroup$
    How good is that , differentiation of $n$ times of this will yield its n-th derivative in terms of n-1th or less.
    $endgroup$
    – M Desmond
    Mar 12 at 15:20












  • $begingroup$
    For $cot x$, see math.stackexchange.com/questions/1701540/…
    $endgroup$
    – lhf
    Mar 13 at 21:21




















  • $begingroup$
    If $y=frac{1}{e^{ax}+b}$, then $y'= a b y^2 - a y$. Perhaps this helps.
    $endgroup$
    – lhf
    Mar 12 at 0:16










  • $begingroup$
    How good is that , differentiation of $n$ times of this will yield its n-th derivative in terms of n-1th or less.
    $endgroup$
    – M Desmond
    Mar 12 at 15:20












  • $begingroup$
    For $cot x$, see math.stackexchange.com/questions/1701540/…
    $endgroup$
    – lhf
    Mar 13 at 21:21


















$begingroup$
If $y=frac{1}{e^{ax}+b}$, then $y'= a b y^2 - a y$. Perhaps this helps.
$endgroup$
– lhf
Mar 12 at 0:16




$begingroup$
If $y=frac{1}{e^{ax}+b}$, then $y'= a b y^2 - a y$. Perhaps this helps.
$endgroup$
– lhf
Mar 12 at 0:16












$begingroup$
How good is that , differentiation of $n$ times of this will yield its n-th derivative in terms of n-1th or less.
$endgroup$
– M Desmond
Mar 12 at 15:20






$begingroup$
How good is that , differentiation of $n$ times of this will yield its n-th derivative in terms of n-1th or less.
$endgroup$
– M Desmond
Mar 12 at 15:20














$begingroup$
For $cot x$, see math.stackexchange.com/questions/1701540/…
$endgroup$
– lhf
Mar 13 at 21:21






$begingroup$
For $cot x$, see math.stackexchange.com/questions/1701540/…
$endgroup$
– lhf
Mar 13 at 21:21












1 Answer
1






active

oldest

votes


















4












$begingroup$

$$dfrac{d^n}{dx^n} f(x) = frac{a^n P_n(e^{ax})}{(e^{ax}+b)^{n+1}}$$
where $P_n$ is a polynomial of degree $n$, with
$P_0(t) = 1$ and
$$ P_{n+1}(t) = t (b + t) P_n'(t) - (n+1) t P(t)$$
I don't think there's a "closed form", unless you count Faà_di_Bruno's formula






share|cite|improve this answer











$endgroup$













  • $begingroup$
    (+1)Thanks sir, can you recommend any references where i can find the proof of Faa di bruno's formula. I have never heard of this before.
    $endgroup$
    – M Desmond
    Mar 12 at 15:16











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

$$dfrac{d^n}{dx^n} f(x) = frac{a^n P_n(e^{ax})}{(e^{ax}+b)^{n+1}}$$
where $P_n$ is a polynomial of degree $n$, with
$P_0(t) = 1$ and
$$ P_{n+1}(t) = t (b + t) P_n'(t) - (n+1) t P(t)$$
I don't think there's a "closed form", unless you count Faà_di_Bruno's formula






share|cite|improve this answer











$endgroup$













  • $begingroup$
    (+1)Thanks sir, can you recommend any references where i can find the proof of Faa di bruno's formula. I have never heard of this before.
    $endgroup$
    – M Desmond
    Mar 12 at 15:16
















4












$begingroup$

$$dfrac{d^n}{dx^n} f(x) = frac{a^n P_n(e^{ax})}{(e^{ax}+b)^{n+1}}$$
where $P_n$ is a polynomial of degree $n$, with
$P_0(t) = 1$ and
$$ P_{n+1}(t) = t (b + t) P_n'(t) - (n+1) t P(t)$$
I don't think there's a "closed form", unless you count Faà_di_Bruno's formula






share|cite|improve this answer











$endgroup$













  • $begingroup$
    (+1)Thanks sir, can you recommend any references where i can find the proof of Faa di bruno's formula. I have never heard of this before.
    $endgroup$
    – M Desmond
    Mar 12 at 15:16














4












4








4





$begingroup$

$$dfrac{d^n}{dx^n} f(x) = frac{a^n P_n(e^{ax})}{(e^{ax}+b)^{n+1}}$$
where $P_n$ is a polynomial of degree $n$, with
$P_0(t) = 1$ and
$$ P_{n+1}(t) = t (b + t) P_n'(t) - (n+1) t P(t)$$
I don't think there's a "closed form", unless you count Faà_di_Bruno's formula






share|cite|improve this answer











$endgroup$



$$dfrac{d^n}{dx^n} f(x) = frac{a^n P_n(e^{ax})}{(e^{ax}+b)^{n+1}}$$
where $P_n$ is a polynomial of degree $n$, with
$P_0(t) = 1$ and
$$ P_{n+1}(t) = t (b + t) P_n'(t) - (n+1) t P(t)$$
I don't think there's a "closed form", unless you count Faà_di_Bruno's formula







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 11 at 19:54









user

5,41411030




5,41411030










answered Mar 11 at 19:47









Robert IsraelRobert Israel

327k23216469




327k23216469












  • $begingroup$
    (+1)Thanks sir, can you recommend any references where i can find the proof of Faa di bruno's formula. I have never heard of this before.
    $endgroup$
    – M Desmond
    Mar 12 at 15:16


















  • $begingroup$
    (+1)Thanks sir, can you recommend any references where i can find the proof of Faa di bruno's formula. I have never heard of this before.
    $endgroup$
    – M Desmond
    Mar 12 at 15:16
















$begingroup$
(+1)Thanks sir, can you recommend any references where i can find the proof of Faa di bruno's formula. I have never heard of this before.
$endgroup$
– M Desmond
Mar 12 at 15:16




$begingroup$
(+1)Thanks sir, can you recommend any references where i can find the proof of Faa di bruno's formula. I have never heard of this before.
$endgroup$
– M Desmond
Mar 12 at 15:16


















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