Last one, I promise.
I can’t find any resource that gives me the identity for the derivative of e raised to the power of X if X itself is subjected to an exponential operation.
Now, we know that where Y=e^X the derivative is Y’=e^X - weird, but true.
And we know that where Y=e^KX, then Y’=Ke^KX.
But what if Y=e^sin(X) or Y=e^(X^2) ?
It’s chain rule, right?
f’r’instance
y=e^x[sup]2[/sup]
then
y’=2xe^x[sup]2[/sup]
and
y"=2x(2xe^x[sup]2[/sup])+2(e^x[sup]2[/sup])
I think you just follow the chain rule. Take the deriviative of the thing in the parentheses, then the next thing and so-on. Then multiply them all together. Here’s a place to start.
Yes, basically:
y = e^f(x)
dy = e^f(x) d(f(x))
which you see as
y = e^x
dy = e^x dx
when f(x) is x.
Therefore,
y=e^sin(x)
dy = cos(x) e^sin(x) dx
to use the example Super Gnat didn’t.
Ordinary chain rule. 
And for
y = e[sup]sin(x)[/sup]
y’=cos(x) e[sup]sin(x)[/sup]
Like everyone else said, it’s the chain rule.
Beee-zarre.
Rickj:
y=e^x, y’=e^x is not weird. It’s not like it’s luck or something. It is defined that way. The natural log is defined as the solution to the the equation: F(x) = F’(x).
I think he meant to day that the exponential is defined by F’ = F. The natural log is defined as the solution to f’(x) = 1/x.
More to the point, one can prove that if f(x) = a^x, then f’(x) is proportional to f(x) and e is defined as the unique number for which the proportionality constant is 1.