Find the Taylors series for f (x)=e2x about 0
Do you mean e^(2x)?
The first derivative is 2 e^2x, and its value at x = 0 is 2.
The second derivative is 4 e^2x, and its value at x=0 is 4.
You should be able to figure out the nth derivative, and apply the Taylor series formula.
If you don't know it, now is a time to learn it.
f(x) = f(0) + x*f'(0) + (x^2/2!)*f''(0) + ... (x^n/n!)*nth derivative of f(0) @ x=0 +...