Question

dy/dx x^lnx
i know there was one just like this but i got it wrong

Answer 1

i got x^ln(x)((1/lnx)+(lnx/x))

Answer 2

\[y=x^{\ln(x)} \]
\[\ln(y)=\ln(x^{\ln(x)})\]
\[\ln(y)=\ln(x) \ln(x)\]
\[\frac{y'}{y}=\frac{1}{x}\ln(x)+\frac{1}{x}\ln(x)\]

Answer 3

\[\Large y = x^{\ln(x)}\]


\[\Large \ln(y) = \ln\left(x^{\ln(x)}\right)\]


\[\Large \ln(y) = \ln(x)\ln(x)\]


\[\Large \ln(y) = (\ln(x))^2\]


\[\Large \frac{d}{dx}[\ln(y)] = \frac{d}{dx}[(\ln(x))^2]\]


\[\Large \frac{y^{\prime}}{y} = \frac{2\ln(x)}{x}\]


\[\Large y^{\prime} = \frac{2y\ln(x)}{x}\]


\[\Large y^{\prime} = \frac{2x^{\ln(x)}\ln(x)}{x}\]

Answer 4

gj jim

Answer 5

thx

Answer 6

so how did you get the 1/x's?

Answer 7

(ln(x))'

Answer 8

=1/x

Answer 9

bingo

Answer 10

either way is find you could use the product rule
or you could multiply (ln(x))(ln(x))=(ln(x))^2
and use chain rule instead

Answer 11

ok i think i get it. maybe after some more work lol ty guys again

Answer 12

i will make my usual plea to understand that
\[f(x)^{g(x)}\] means
\[e^{g(x)\ln(f(x))}\] pretty much by definition. the work for finding the derivative is identical. you have to find the derivative of
\[g(x)\ln(f(x))\] but conceptually it seems less like magic and more like the definition of exponentiation.