Find the first and second derivate of  y = e^9e^x?

y = e^9e^x

y = e^(x + 9)

dy/dx = e^(x + 9)

for 2nd derivative

d^2y/dx^2 = e^(x + 9) Answer//

y=(e^9)e^x

=>

y'=(e^9)e^x

=>

y"=(e^9)e^x~8103.08e^x

y = e^9e^x

y = e^(x + 9)

If this is (e^9) * (e^x), then it's simple.  e^9 is just a constant.  Call it k (or anything, it doesn't matter).

y = k * e^x

y' = k * e^x

y'' = k * e^x

y''' = k * e^x

And so on, forever.

y = (e^9) * (e^x)

y' = (e^9) * (e^x)

y'' = (e^9) * (e^x)

If it's e^(9 * e^(x)), then that's a whole new ball of wax

y = e^(9 * e^(x))

ln(y) = 9 * e^(x)

y' / y = 9 * e^(x)

y' = 9 * e^(x) * y

y' = 9 * e^(x) * e^(9 * e^(x))

y' = 9 * e^(x + 9 * e^(x))

y' = e^ln(9) * e^(x + 9 * e^(x))

y' = e^(x + 9 * e^(x) + ln(9))

ln(y') = x + 9 * e^(x) + ln(9)

y'' / y' = 1 + 9 * e^(x) + 0

y'' = y' * (1 + 9 * e^(x))

y'' = 9 * e^(x + 9 * e^(x)) * (1 + 9 * e^(x))

Presuming that is:

y = e⁹ e^x

This simplifies to:

y = e^(x + 9)

The first derivative needs the chain rule:

y = e^u and u = x + 9

dy/du = e^u and du/dx = 1

dy/dx = dy/du * du/dx

dy/dx = e^u * 1

dy/dx = e^u

dy/dx = e^(x + 9)

So:

y = e^(x + 9)

y' = e^(x + 9)

Therefore:

y" = e^(x + 9)

If you meant something else you'll have to be more specific as to what you meant with proper use of parenthesis.

First of all, type that out correctly, with parentheses, so we can figure out what in the world you are talking about.

This will probably involve multiple applications of the chain rule. And remember that the derivative of e^x is e^x. That's something that you need to memorize.