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Francis Coffey
Mathematics
23 August, 02:29
If y=u + 2e^u and u=1+lynx
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Larissa Coffey
23 August, 02:32
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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
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Lily Combs
23 August, 02:38
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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
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