$\displaystyle \frac{e^t-1}{t}$

I know this is a simple derivative, and I'm in calc 3, but I can't seem to figure this one out....

Okay, well that's really small. It's e^t-1/t in case you can't see it.

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- Mar 23rd 2011, 11:39 AMBracketologySimple derivative...
$\displaystyle \frac{e^t-1}{t}$

I know this is a simple derivative, and I'm in calc 3, but I can't seem to figure this one out....

Okay, well that's really small. It's e^t-1/t in case you can't see it. - Mar 23rd 2011, 11:47 AMQuacky
Have you tried using the quotient rule?

Let $\displaystyle f(t)=e^t-1$

Then, $\displaystyle f'(t)=e^t$

Let $\displaystyle g(t)=t$

$\displaystyle g'(t)=1$

The rule states that:

For any function composed of a quotient of two other functions, here denoted by $\displaystyle \displaystyle\frac{f(t)}{g(t)}$, the derivative is:$\displaystyle \displaystyle\frac{f'(t)g(t)-g'(t)f(t)}{[g(t)]^2}$

...So do the substitution.

If you haven't covered the quotient rule, then reply and I'll try another approach (Wink) - Mar 23rd 2011, 01:40 PMBracketology
Yea, I'm in calc 3....I was just having a brain fart. Thanks. >.<