Question

derivate x^2x

Answer 1

\[y=x ^{2x} \]
\[\ln y=\ln x ^{2x}\]
\[\ln y=2x.\ln x\]
\[(\ln y)\prime=(2x.\ln x)\prime\]
\[y \prime/y=2.\ln x+1/x.2x\]
\[y \prime=(2.\ln x+2).y\]
\[y \prime=2.y.(\ln x+1)\]
\[y \prime=2.x ^{2x}.(\ln x+1)\]

Answer 2

Let u = 2x. then du /dx = 2
then dy/dx = dy/du/du/dx = u *x^(u-1) du/dx = 2(2x)x^(2x-1)= 4x^2x)

Answer 3

1st answer might be correct....2nd answer i dont think so because it was in my test worth 6marks....so i cant get 6 marks for such simple stuff there is a trick to this queestion

Answer 4

You're right. There is a trick.
f = x^2x
we know that a^x = e^(e^xlna) let u = e^2xlnx; du/dx = (e^2x)/x + (2e^2x)ln x
so f' =dy/du)(du/dx)= [e^(xlna)] (ln a) = a^x lna} du/dx
for our problem it would be (x^2x ln 2x)[(e^2x)/x + 2e^2x)ln x)
I think.

Answer 5

tad is wrong the easiest way to do so is to do this
\[e^(2xlnx)=e^(lny) \]
then differinitiate the left side only
e^2xlnx(2lnx+2)

then you get x^2x(2ln2x+2)=dy/dx because the e^ln cancels you just get y prime

hope this helps