Find dy/dx for the function xy^2 +x lnx =4y for x>0?

Question

please

Answer 1

xy^2 +x lnx =4y

apply differential to both sides:

d[xy^2 +x lnx] = d [4y]

d[xy^2] + d[x lnx] = d [4y]

d[x] y^2 + x d[y^2]    +    d[x] lnx + x dln[x] = 4d[y]

y^2 dx + x 2ydy    +     lnx dx+ x 1/x dx = 4dy

group terms with dx and dy:

(y^2 + lnx + x/x) dx +  2xy dy = 4dy

(y^2 + lnx + 1) dx =  (4 - 2xy)dy

finish by yourself

Answer 2

u hv to use chain ruke in this case,since v r differentiating wrt X in case of y we wil treat it as X bt will then also multiply it wid dy/dx......

differentiating d given expression wrt X,

2x[der of x^n= n. x^n-1]  y^2 + x^2 2y dy/dx[derivate y as x only ] +  log x . 1[since  derivative of x is 1]+x . 1/x[since derivative of log x is 1/x]=4 dy/dx[der of y=1 bt since v r derivating it wrt to x multipy it wid dy/dx]

nw take dy/dx on one side. ur expression nw bcumz as

[2xy^2 +logx +1] / 4-x^2y