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

The d/(dn)(n^n) is n^n(ln(n)+1)

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

The first d/(dn)(n^n) is n^n(ln(n)+1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

Solution steps

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

= e^{n l n (n)} (ln (n) + 1)

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

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