91Ó°ÊÓ

The given function has two terms: 1. \(\sin x\) 2. \(4e^{0.5x}\)

We will differentiate the first term, \(\sin x\), with respect to \(x\). The derivative of \(\sin x\) with respect to \(x\) is \(\cos x\). So, we have: $$\frac{d(\sin x)}{dx} = \cos x$$

Now we will differentiate the second term, \(4e^{0.5x}\), with respect to \(x\). To do this, we will apply the chain rule since the term is a composite function. The chain rule states that if we have a function \(u(v(x))\), then the derivative of \(u\) with respect to \(x\) is: $$\frac{du}{dx} = \frac{du}{dv}\cdot\frac{dv}{dx}$$ In our case, we have a function of the form \(4e^{0.5x}\), which can be written as \(u(v(x))\), where \(u(v) = 4e^v\) and \(v(x) = 0.5x\). Now we need to find the derivative of \(u\) with respect to \(v\) and the derivative of \(v\) with respect to \(x\): 1. \(\frac{du}{dv} = \frac{d(4e^v)}{dv} = 4e^v\) 2. \(\frac{dv}{dx} = \frac{d(0.5x)}{dx} = 0.5\) Applying the chain rule now, we get the derivative of the second term: $$\frac{d(4e^{0.5x})}{dx} = 4e^v \cdot 0.5 = 2e^{0.5x}$$

Finally, we will combine the derivatives of both terms to get the derivative of the whole function \(y\). Since the function \(y\) is the sum of the two terms, according to the sum rule of differentiation, the derivative of \(y\) with respect to \(x\) is the sum of the derivatives of the individual terms. Therefore, we have: $$\frac{dy}{dx} = \frac{d(\sin x)}{dx} + \frac{d(4e^{0.5x})}{dx} = \cos x + 2e^{0.5x}$$ Now we have found the derivative of the function \(y\) with respect to \(x\): $$\frac{dy}{dx} = \cos x + 2e^{0.5x}$$