Question

Can someone please explain how to differentiate y=2x^x i keep getting it wrong. I thought you do xln2x= dy/dx 1/y

Answer 1

Did you use logarithmic differentiation? It seems like it from your steps, although it's a bit hard to read it that way:
\[\Large \ln(y(x))=x\ln(2x) \]

Answer 2

i did, what do you do from there

Answer 3

Differentiate, notice that y is a function of x, therefore apply the chain rule:

\[\Large \frac{1}{y(x)}\frac{dy}{dx}=\ln(2x)+\frac{1}{2x}\cdot2\cdot x \]

Answer 4

can any of u help with a question?

Answer 5

so it goes to ln2 +xlnx= (0+lnx+1)*2x^x