Question

differentiate x^x

Answer 1

\[x^{x}\]

Answer 2

don't think that's right yahoo.

Answer 3

let \[y = x^x\]
take ln of both sides \[\ln y = x\ln x\]
implictly differentiate
\[\frac{1}{y} y' = x(\frac 1x) + \ln x (1)\]
\[\frac 1y y' = 1 + \ln x\]
\[y = y(1 + \ln x)\]
\[y = x^x (1 + \ln x\]

Answer 4

got it?

Answer 5

Dafuq was that?
@lgbasallote

Answer 6

ok
let
y=x^x
take ln of both sides
ln(y)=ln(x^x)
ln(y)=x(ln(x)
differentiate both sides (use product and chain rule)

(1/y)y'=(ln(x)+1)
or multiply by y
y'=y(ln(x)+1


can you tell me the answer now ?

Answer 7

thanks :)

Answer 8

Can you also explain, \[x ^{x ^{x}}\]

Answer 9

@Beven just take a look at the above procedure is same for this one .

Answer 10

so ln(y)=ln(x^x^x)

Answer 11

is ln(x^x^x) just x.ln(x^x)?

Answer 12

no it is x^xln(x)

Answer 13

ah okay.

Answer 14

thanks :)