derivative?!?

Question

please explain and solve problem step by step. thank you!

if y=x^2 + y lnx , find dy/dx

Answer 1

Find the derivative implicitly (I will use y' for dy/dx):

y' = 2x + y/x + lnx*y'

Gather any terms with y' to one side:

y' - lnx * y' = 2x + y/x

Factor out y' and then solve for y':
y' (1 - lnx) = 2x + y/x

y' = (2x + y/x)/(1-lnx)

y' = (2x² + y)/(x-xlnx)

Answer 2

You have to differentiate implicitely

dy/dx = 2x + y(1/x) + dy/dx * lnx
dy/dx ( 1 - lnx) = 2x + y/x
dy/dx = (2x + y/x) / (1 - lnx)

Answer 3

y - ylnx = x^2
y(1 - lnx) = x^2
y = x^2 / (1 - lnx)

dy / dx = [2x(1 - lnx) + x] / (1 - lnx)²

Answer 4

First you have to isolate the y. So,

y - y*ln(x) = x^2
y*(1-ln(x)) = x^2

So, divide the (1-ln(x)) over, and we have a function to differentiate:

y = x^2/(1-ln(x)).

Now, you have to do the quotient rule, but it shouldn't be too bad. Go from here, as this is probably what got you tied up.

Answer 5

y' = 2x y'lnx + y/x
y' -2xlnxy' = y/x
y'(1-2xlnx) = y/x
y' = y/[x(1-2xlnx)]