let $f(x)=2e^{x}-5cos(x)+3$. $f(x)=$

Explanation:

Step1: Differentiate (2e^x)

The derivative of (e^x) is (e^x). Using the constant multiple rule, the derivative of (2e^x) is (2e^x).

Step2: Differentiate (-5\cos(x))

The derivative of (\cos(x)) is (-\sin(x)). Using the constant multiple rule, the derivative of (-5\cos(x)) is (5\sin(x)).

Step3: Differentiate (3)

The derivative of a constant (C) (here (C = 3)) is (0).

Step4: Sum up the derivatives

By the sum rule of differentiation ((u + v+w)'=u'+v'+w'), where (u = 2e^x), (v=-5\cos(x)) and (w = 3). So (f'(x)=2e^x+5\sin(x)+0).

Answer:

(2e^x + 5\sin(x))