What is the d/(dn)((s-n)ln(1-p)) ?

The d/(dn)((s-n)ln(1-p)) is -ln(1-p)-(\frac{dp)/(dn)(s-n)}{1-p}

What is the first d/(dn)((s-n)ln(1-p)) ?

The first d/(dn)((s-n)ln(1-p)) is -ln(1-p)-(\frac{dp)/(dn)(s-n)}{1-p}

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\frac{d}{d n} ((s - n) ln (1 - p))

- ln (1 - p) - \frac{\frac{d p}{d n} ( s - n )}{1 - p}

Solution steps

\frac{d}{d n} ((s - n) ln (1 - p))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = s - n, g = ln (1 - p)

= \frac{d}{d n} (s - n) ln (1 - p) + \frac{d}{d n} (ln (1 - p)) (s - n)

\frac{d}{d n} (s - n) = - 1

\frac{d}{d n} (ln (1 - p)) = - \frac{\frac{d p}{d n}}{1 - p}

= (- 1) ln (1 - p) + (- \frac{\frac{d p}{d n}}{1 - p}) (s - n)

Simplify

(- 1) ln (1 - p) + (- \frac{\frac{d p}{d n}}{1 - p}) (s - n) : - ln (1 - p) - \frac{\frac{d p}{d n} ( s - n )}{1 - p}

= - ln (1 - p) - \frac{\frac{d p}{d n} ( s - n )}{1 - p}

1 5^{10} \cdot \frac{1}{10 !} \cdot e^{- 15}

2000 e^{- 0.06 (3)}

π^{- \frac{1}{4}}

- 2.18 \cdot 1 0^{- 18} (\frac{1}{1 ^{2}} - \frac{1}{3 ^{2}})

What is the d/(dn)((s-n)ln(1-p)) ?
The d/(dn)((s-n)ln(1-p)) is -ln(1-p)-(\frac{dp)/(dn)(s-n)}{1-p}
What is the first d/(dn)((s-n)ln(1-p)) ?
The first d/(dn)((s-n)ln(1-p)) is -ln(1-p)-(\frac{dp)/(dn)(s-n)}{1-p}