Question

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:

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