What is the d/(dn)((1-i)^{n-1}) ?

The d/(dn)((1-i)^{n-1}) is ln(1-i)(1-i)^{n-1}

What is the first d/(dn)((1-i)^{n-1}) ?

The first d/(dn)((1-i)^{n-1}) is ln(1-i)(1-i)^{n-1}

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\frac{d}{d n} ((1 - i)^{n - 1})

ln (1 - i) (1 - i)^{n - 1}

Solution steps

\frac{d}{d n} ((1 - i)^{n - 1})

Apply exponent rule:

a^{b} = e^{b l n (a)}

(1 - i)^{n - 1} = e^{(n - 1) l n (1 - i)}

= \frac{d}{d n} (e^{(n - 1) l n (1 - i)})

Apply the chain rule:

e^{(n - 1) l n (1 - i)} \frac{d}{d n} ((n - 1) ln (1 - i))

= e^{(n - 1) l n (1 - i)} \frac{d}{d n} ((n - 1) ln (1 - i))

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

= e^{(n - 1) l n (1 - i)} ln (1 - i)

e^{(n - 1) l n (1 - i)} = (1 - i)^{n - 1}

= ln (1 - i) (1 - i)^{n - 1}

What is the d/(dn)((1-i)^{n-1}) ?
The d/(dn)((1-i)^{n-1}) is ln(1-i)(1-i)^{n-1}
What is the first d/(dn)((1-i)^{n-1}) ?
The first d/(dn)((1-i)^{n-1}) is ln(1-i)(1-i)^{n-1}