Does 2e^5X differentiated = 10e^5X

Also how would I integrate 3e^-x?

Using the chain rule;

d/dx of 2e^5x = 2e^5x(d/dx 5x) = 2e^5x(5) = 10e^5x

to integrate 3e^-x first you must know the following;

the derivative of 3e^-x = 3e^-x(d/dx -x) = 3e^-x(-1) = -3e^-x

Hope this helps.

Yes, you're right.
f(x) = 2e^(5x)
f'(x) = 2e^(5x) * 5
f'(x) = 10e^(5x)

If you differentiate e^x it stays e^x, in this case you also need to apply the chain rule.

Yes, 2e^5X differentiated = 10e^5X.
d/dx (3e^-x) = -3e^-x

You need to use the chain rule

d/dx [2e^5x] = 2e^5x d/dx [5x] = 2e^5x (5) = 10e^5x

integral of 3e^-x = - 3e^-x