en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

d/(dn)((2^n)/(n!))

Solution

\frac{d}{d n} (\frac{2 ^{n}}{n !})

Solution

\frac{2 ^{n} ln ( 2 ) n ! - Γ ( n + 1 ) ϝ ( n + 1 ) \cdot 2 ^{n}}{n ! ^{2}}

Solution steps

\frac{d}{d n} (\frac{2 ^{n}}{n !})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d n} ( 2 ^{n} ) n ! - \frac{d}{d n} ( n ! ) 2 ^{n}}{n ! ^{2}}

\frac{d}{d n} (2^{n}) = 2^{n} ln (2)

\frac{d}{d n} (n!) = Γ (n + 1) ϝ (n + 1)

= \frac{2 ^{n} ln ( 2 ) n ! - Γ ( n + 1 ) ϝ ( n + 1 ) \cdot 2 ^{n}}{n ! ^{2}}

Graph

Popular Examples

derivative of y=-3^2+x+2 derivative of

y = - 3^{2} + x + 2

derivative of (xy-1^2)

\frac{d}{d x} ((x y - 1)^{2})

derivative of ln((2x-3)^2)derivative of

ln ((2 x - 3)^{2})

derivative of (1/x)(x^2-2)derivative of

(\frac{1}{x}) (x^{2} - 2)

derivative of sin(x^{1/3})derivative of

sin (x^{\frac{1}{3}})

Frequently Asked Questions (FAQ)

What is the d/(dn)((2^n)/(n!)) ?
The d/(dn)((2^n)/(n!)) is (2^nln(2)n!-\Gamma(n+1)\digamma(n+1)*2^n)/(n!^2)
What is the first d/(dn)((2^n)/(n!)) ?
The first d/(dn)((2^n)/(n!)) is (2^nln(2)n!-\Gamma(n+1)\digamma(n+1)*2^n)/(n!^2)