What is the d/(dn)(i^{2n}) ?

The d/(dn)(i^{2n}) is 2ln(i)i^{2n}

What is the first d/(dn)(i^{2n}) ?

The first d/(dn)(i^{2n}) is 2ln(i)i^{2n}

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

2 ln (i) i^{2 n}

Solution steps

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

Apply exponent rule:

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

i^{2 n} = e^{2 n l n (i)}

= \frac{d}{d n} (e^{2 n l n (i)})

Apply the chain rule:

e^{2 n l n (i)} \frac{d}{d n} (2 n ln (i))

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

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

= e^{2 n l n (i)} \cdot 2 ln (i)

e^{2 n l n (i)} = i^{2 n}

= 2 ln (i) i^{2 n}

cot (x^{4})

(x + 2) (x^{3} - 1)

\frac{d}{dd} (d + 14)

x^{2} (1 - m^{2})

f (x) = 10 x^{2} + 5 x - 2