What is the d/(dn)((n!)/(7^n)) ?

The d/(dn)((n!)/(7^n)) is (\Gamma(1+n)\digamma(1+n)-ln(7)n!)/(7^n)

What is the first d/(dn)((n!)/(7^n)) ?

The first d/(dn)((n!)/(7^n)) is (\Gamma(1+n)\digamma(1+n)-ln(7)n!)/(7^n)

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\frac{d}{d n} (\frac{n !}{7 ^{n}})

\frac{Γ ( 1 + n ) ϝ ( 1 + n ) - ln ( 7 ) n !}{7 ^{n}}

Solution steps

\frac{d}{d n} (\frac{n !}{7 ^{n}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d n} ( n ! ) 7 ^{n} - \frac{d}{d n} ( 7 ^{n} ) n !}{( 7 ^{n} ) ^{2}}

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

\frac{d}{d n} (7^{n}) = 7^{n} ln (7)

= \frac{Γ ( n + 1 ) ϝ ( n + 1 ) \cdot 7 ^{n} - 7 ^{n} ln ( 7 ) n !}{( 7 ^{n} ) ^{2}}

Simplify

\frac{Γ ( n + 1 ) ϝ ( n + 1 ) \cdot 7 ^{n} - 7 ^{n} ln ( 7 ) n !}{( 7 ^{n} ) ^{2}} : \frac{Γ ( 1 + n ) ϝ ( 1 + n ) - ln ( 7 ) n !}{7 ^{n}}

= \frac{Γ ( 1 + n ) ϝ ( 1 + n ) - ln ( 7 ) n !}{7 ^{n}}

1 + x tan (x)

3 x^{2} - 5 x^{3} + 4 x^{20}

y = e^{4} + ln (x)

3 sin^{2} (x) + 2 tan (x)

(2 x^{2} + 3 x + 2)^{5}

What is the d/(dn)((n!)/(7^n)) ?
The d/(dn)((n!)/(7^n)) is (\Gamma(1+n)\digamma(1+n)-ln(7)n!)/(7^n)
What is the first d/(dn)((n!)/(7^n)) ?
The first d/(dn)((n!)/(7^n)) is (\Gamma(1+n)\digamma(1+n)-ln(7)n!)/(7^n)