What is the derivative of e^x/(1+e^y)ex1+ey?

1 Answer
Aug 30, 2015

d/dx(e^x/(1+e^y))= (e^x(1+e^y-e^ydy/dx))/(1+e^y)^2ddx(ex1+ey)=ex(1+eyeydydx)(1+ey)2

d/dt(e^x/(1+e^y))= (e^x(dx/dt+dx/dt e^y-e^ydy/dt))/(1+e^y)^2ddt(ex1+ey)=ex(dxdt+dxdteyeydydt)(1+ey)2

Explanation:

Assuming that we are differentiation with respect to xx and that yy is some function of xx, we can differentiate implicitly using the quotient rule:

d/dx(e^x/(1+e^y)) = (e^x(1+e^y)-e^x(e^y dy/dx))/(1+e^y)^2ddx(ex1+ey)=ex(1+ey)ex(eydydx)(1+ey)2

If we are differentiating with respect to some tt and assuming that xx and yy are functions of tt, we get:

d/dt(e^x/(1+e^y)) = (e^x dx/dt (1+e^y)-e^x(e^y dy/dt))/(1+e^y)^2ddt(ex1+ey)=exdxdt(1+ey)ex(eydydt)(1+ey)2

= (e^x[dx/dt(1+e^y)-e^x(e^y dy/dt)])/(1+e^y)^2=ex[dxdt(1+ey)ex(eydydt)](1+ey)2

= (e^x(dx/dt+dx/dt e^y-e^ydy/dt))/(1+e^y)^2=ex(dxdt+dxdteyeydydt)(1+ey)2