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Popular >

d/(dn)(i^{4n+1})

Solution

\frac{d}{d n} (i^{4 n + 1})

Solution

4 ln (i) i^{4 n + 1}

Solution steps

\frac{d}{d n} (i^{4 n + 1})

Apply exponent rule:

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

i^{4 n + 1} = e^{(4 n + 1) l n (i)}

= \frac{d}{d n} (e^{(4 n + 1) l n (i)})

Apply the chain rule:

e^{(4 n + 1) l n (i)} \frac{d}{d n} ((4 n + 1) ln (i))

= e^{(4 n + 1) l n (i)} \frac{d}{d n} ((4 n + 1) ln (i))

\frac{d}{d n} ((4 n + 1) ln (i)) = 4 ln (i)

= e^{(4 n + 1) l n (i)} \cdot 4 ln (i)

e^{(4 n + 1) l n (i)} = i^{4 n + 1}

= 4 ln (i) i^{4 n + 1}

Graph

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Frequently Asked Questions (FAQ)

What is the d/(dn)(i^{4n+1}) ?
The d/(dn)(i^{4n+1}) is 4ln(i)i^{4n+1}
What is the first d/(dn)(i^{4n+1}) ?
The first d/(dn)(i^{4n+1}) is 4ln(i)i^{4n+1}