# Second Derivative of e^t + t^e ?

## Homework Statement

Find the second derivative of e^t + t^e :

## Homework Equations

(e^u) = u' * (e^u)

## The Attempt at a Solution

e^t + t^e

1* (e^t) + e*[t^(e-1)]

^^^first derivative

(e^t) + e*e[t^{(e-1)-1}]

^^^ 2nd derivative

Answer?: (e^t) + e^2[t^(e-2)]

## Answers and Replies

Answer?: (e^t) + e^2[t^(e-2)]

you're quite close. after taking the first derivative, the exponent of t changes, you have not accounted for that.

I'm not sure I follow. Doesn't the exponent of t become (e-2) ?

d/dt (t^e) = e*t^(e-1)

d/dt (e*t^(e-1)) = ??

EDIT/HINT: Your exponent is correct, it is the constant in front of the exponent that isn't. It's quite a common error, though you probably won't be making it in the future.

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scratch that... it didn't even look almost right

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e^t + t^e

1* (e^t) + e*[t^(e-1)]

$$t^e$$
$$t^u = u' f'(u)$$

What is u'?

u' = e

?

:/

u' = e

?

:/

Wrong, derivative of e^x is e^x; don't confuse that with "e"

hint: e itself is equal to what?