Find dy/dx when y = 2ln(2e-x)

Answering this question requires knowledge of e and ln. If we look at the question we can see 2 is being multiplied by the ln, so we might want to use the product rule. Also, we can see that the ln has a big term in side of it, so we might want to use the chain rule to simplify this down. For a general case, we know if y=ln(x) then dy/dx = 1/x. So we can now use the chain rule on ln(2e-x). If we let u = 2e-x, we have that du/dx = -1 (remember that 2e is just a constant so when you differentiate it you get 0). We now have y = ln(u) so dy/du = 1/u. Applying the chain rule formula gets us dy/dx = dy/du * du/dx = 1/(2e-x) * -1 = -1/(2e-x). However, we are not yet done as ln is multiplied by 2. Since 2 is just a multiplying constant to the ln, we can use the product rule. Letting u = 2 and v = ln(2e-x) we know du/dx = 0 and dv/dx = -1/2e-x, meaning for y=2ln(2e-x), we have that dy/dx = udv/dx + vdu/dx = 2* -1/(2e-x) + 0 = -2/(2e-x).