Question

differentiate the following function.
$y = 2e^{x}+3x$
$\frac{dy}{dx}=\square$

Explanation:

Step1: Differentiate \(2e^x\)

The derivative of \(e^x\) is \(e^x\). Using the constant - multiple rule \((cf(x))^\prime = cf^\prime(x)\), for \(y_1 = 2e^x\), we have \(y_1^\prime=2\frac{d}{dx}(e^x)=2e^x\).

Step2: Differentiate \(3x\)

Using the power rule \(\frac{d}{dx}(x^n)=nx^{n - 1}\), for \(y_2 = 3x\) (where \(n = 1\)), we have \(y_2^\prime=3\frac{d}{dx}(x)=3\times1\times x^{1-1}=3\).

Step3: Use the sum rule

The sum rule states that \((u + v)^\prime=u^\prime+v^\prime\). If \(y = 2e^x+3x\) (where \(u = 2e^x\) and \(v = 3x\)), then \(\frac{dy}{dx}=\frac{d}{dx}(2e^x)+\frac{d}{dx}(3x)\).

Answer:

\(2e^x + 3\)