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)

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\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)

n = 0 \sum \infty n! (x - 5)^{n}

\int \frac{1}{x ( 1 + ln ( x ) ) ^{3}} d x

\frac{d r}{d θ} = \frac{r ^{2} + 2 r θ}{θ ^{2}}

x \to 0 lim (\frac{x}{sin ( 8 x )})

x \to 1 lim (e^{x - 1} + 1)