find $y^{\text{'}}.$

$x^2+e^{xy}+y^2=1$

$\frac{2x^{\text{'}}}{(1+x^4)e^{y-x}}.$

$\frac{{?}^{2x}}{lnx}\frac{.a}{(1+x^4)e^y}$

none

$\frac{2x+y{e}^{xy}}{2y+x{e}^{xy}}$

Given the information provided in Fig. $4\mathrm{.}126,$ determine:

a$.I_C.$

b$.V_E.$

c$.V_B.$

d$.R_1.$

FIG. $4\mathrm{.}125$

Problems $15,16,20,23,25,67.69,70,73,$ and $77.$

FIG. $4\mathrm{.}126$