Question

If y=u +2e^u and u=1+lynx

Answer 1

i am not very good at math but i wanna say this is how you do it not a 100% sure

y = u + 2e^u
dy/du = 1 + 2e^u

u = 1 + ln x
du/dx = 1/x

Then dy/dx = dy/du * du/dx

dy/dx = (1+2e^u) * 1/x

Now replace u with 1 + ln x

dy/dx = (1 + 2e^(1+lnx)) * 1/x
dy/dx = (1 + 2e^1 * e^ln x) * 1/x
dy/dx = (1 + 2e x) * 1/x
dy/dx = 1/x + 2e

Now let x = 1/e. Then

dy/dx = 1/(1/e) + 2e = e + 2e = 3e

Answer 2

y = u + 2e^u
dy/du = 1 + 2e^u

u = 1 + ln x
du/dx = 1/x

Then dy/dx = dy/du * du/dx

dy/dx = (1+2e^u) * 1/x

Now replace u with 1 + ln x

dy/dx = (1 + 2e^(1+lnx)) * 1/x
dy/dx = (1 + 2e^1 * e^ln x) * 1/x
dy/dx = (1 + 2e x) * 1/x
dy/dx = 1/x + 2e

Now let x = 1/e. Then

dy/dx = 1/(1/e) + 2e = e + 2e = 3e