To solve the problem of partial derivatives in higher mathematics, how did the second question get this answer? What does u(x, 2x) mean?

U(x, y) is a binary function, and u(x, 2x) is a univariate function about x obtained by substituting y = 2x into it.

Similarly, the partial derivatives u'x(x, y) and u"xx(x, y) are both binary functions. Substituting y = 2x can get u'x(x, 2x) and u"xx(x, 2x).

According to the chain derivative law, the derivative of unary function u(x, 2x) is u' x (x, 2x) 1+u' y (x, 2x) 2 = u' x (x, 2x)+2u' y (x, 2x).

∫u(x, 2x) = x and u'x(x, 2x) = x? ，∴u'y(x,2x) = ( 1-x？ )/2.

Furthermore, the derivative of the unary function u'x(x, 2x) is u "xx (x, 2x) 1+u" yx (x, 2x) 2 = u "xx (x, 2x)+2u" yx (x, 2x).

∫u ' x(x，2x) = x？，∴2x = u”xx(x，2x)+2u”yx(x，2x) ①。

Similar to -x = (( 1-x? )/2) = (u'y(x，2x))' = u"xy(x，2x)+2u"yy(x，2x) ②。

The second partial derivative of ∵u is continuous, ∴ u "xy (x, y) = u" yx (x, y).

Then for ①-②, 4x = u"xx(x, 2x)-4u"yy(x, 2x).

Equation u"xx-u"yy = 0 is satisfied by u and substituted into -3u"xx(x, 2x) = 4x, that is, u"xx(x, 2x) = -4x/3.