# Thread: trouble taking a derivative

1. ## trouble taking a derivative

I need help taking the following derivatives:

d/dt of (e^-2)*(e^(1+e^x))*(e^t)

and

d/dt of (e^-2)*(e^(1+e^t))*(e^x)

thanks

Fred1956

2. Originally Posted by Fred1956
I need help taking the following derivatives:

d/dt of (e^-2)*(e^(1+e^x))*(e^t)
If $\displaystyle x \neq f(t)$ let $\displaystyle u = e^{-2}(e^{1+e^x})$ which is a constant which gives:

$\displaystyle \frac{d}{dt}(ue^t)$ where u is a constant, should be easy enough.

d/dt of (e^-2)*(e^(1+e^t))*(e^x)
$\displaystyle \frac{d}{dt}(e^{-2}e^x)(e^{1+e^t})$

Let $\displaystyle y = (e^{-2}e^x)(e^{1+e^t})$

take logs:

Spoiler:
$\displaystyle ln(y) = ln[(e^{-2}e^x)(e^{1+e^t})]$

(Because $\displaystyle ln(abc) = ln(a)+ln(b)+ln(c)$)
$\displaystyle = ln(e^{-2}) + ln(e^x) + ln(e^{1+e^t})$

(Because $\displaystyle ln(a^k) = k\,ln(a)$ and $\displaystyle ln(e^k) = k$)
$\displaystyle = -2 + x + 1+e^t = e^t+x-1$

$\displaystyle ln(y) = e^t+x-1$

Differentiate implicitly:

$\displaystyle \frac{1}{y} \frac{dy}{dt} = e^t$

$\displaystyle \frac{dy}{dt} = ye^t = e^t(e^{-2}e^x)(e^{1+e^t})$