how i do this? (1 Viewer)

[bos1234]

Differentiate, either by expressing as a power of e, or by taking logs of boths sides and using implicit differentiation
y=

Please show both methods if possible

 

[ssglain]

This might get a bit confusing using plain text..

Method 1 requires no more than 2/3U skills:
Remember a = e^ln(a) and ln(b^c) = c.ln(b) so:
y = x^lnx = e^ln[x^(lnx)] = e^[(lnx).(lnx)] = e^[(lnx)^2]

Now you have y = e^f(x)
f(x) = [ln(x)]^2
f'(x) = 2ln(x).(1/x)

dy/dx = f'(x).[e^f(x)] = 2(lnx).(1/x).{e^[(lnx)^2]} = 2(lnx).(1/x).(x^lnx)
Therefore, dy/dx = [2(lnx).(x^lnx)]/x

Method 2 uses 4U implicit differentiation:
y = x^lnx
ln(y) = ln[x^(lnx)] = (lnx).(lnx) = (lnx)^2
(1/y)(dy/dx) = 2(lnx).(1/x)
dy/dx = 2y.(lnx).(1/x) = 2(x^lnx).(lnx).(1/x)
Therefore, dy/dx = [2(lnx).(x^lnx)]/x