Calculus involving exponents

1) how do you differentiate this: Ae^uθ

2) how do you integrate this: 50(2 - 3e^-t/2)

I have made a few attempts but I'm not sure if I got it right, any help would be awesome
off to youtube to learn more calculus! =D

exponentials are tough

use the chain rule

d(e^(uX))/dX = d(e^(uX))/d(uX) * d(uX)/dX

now its easy

for my example the answer is

= e^(uX) * u = u*e^(uX)

so you can see take the deritave of the power and then put it in front of the same exponential value as you started with

so your question i will use X instead of theta

1) d(Ae^uX)/dX = d(uX)/dX * Ae^uX = A*u*e^(uX)

ok see what i did

the A is a constant so it can go outside the derivative and can be multiplied back in later

2)integral 50(2-3e^(-t/2)) dt

im assuming the dt since you said integrate

first the 50 is constant it can go outside

= 50 * int (2-3e^(-t/2)) dt
= 50 * (int 2 dt - int 3*e^(-t/2)) dt )

im thinking that e^(-t/2) will produce -0.5*e^(-t/2) if we take the derivative
so if we have A*e^(-t/2) and its derivative is -0.5*A*e^(-t/2) then the first value A*e^(-t/2) must be the anti derivative of -0.5*A*e^(-t/2)


since we want the anti derivative of
3*e^(-t/2) = -0.5*A*e^(-t/2)
3 = -0.5*A
A = -6

this means that -6*e^(-t/2) is the anti derivative of 3*e^(-t/2)

we can confirm this by taking the derivative

-6*-1/2*e^(-t/2) = 3*e^(-t/2) so its true
back to the question

= 50 * (int 2 dt - int 3*e^(-t/2)) dt )
= 50 * 2t - 6*e^(-t/2) + C
that should be correct

1/
Do you differentiate with respect to u or θ ???

2/
= 50(2t + 6e^(-t/2)) + c