Finding the Derivative of \(y = e^x \sin x\)

The problem asks us to find the derivative of the function $y = e^x \sin x$ with respect to $x$. This means we need to calculate $\frac{dy}{dx}$.

Understanding the Product Rule

To find the derivative of a function that is the product of two other functions, we use the Product Rule for differentiation. The rule states that if we have a function $y = u(x) \cdot v(x)$, then its derivative is given by:

\(\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)\)

Here, $u'(x)$ is the derivative of $u(x)$ and $v'(x)$ is the derivative of $v(x)$.

Applying the Product Rule

Let's identify the two functions in our problem, $y = e^x \sin x$:

• Let $u(x) = e^x$
• Let $v(x) = \sin x$

Now, we need to find the derivatives of $u(x)$ and $v(x)$:

• The derivative of $u(x) = e^x$ is $u'(x) = \frac{d}{dx}(e^x) = e^x$.
• The derivative of $v(x) = \sin x$ is $v'(x) = \frac{d}{dx}(\sin x) = \cos x$.

Substitute these into the product rule formula:

\(\frac{dy}{dx} = (e^x)(\sin x) + (e^x)(\cos x)\)

Simplifying the Result

We can simplify the expression by factoring out the common term $e^x$:

\(\frac{dy}{dx} = e^x \sin x + e^x \cos x\)

This can also be written as:

\(\frac{dy}{dx} = e^x (\sin x + \cos x)\)

Matching with the Options

Comparing our result, $e^x \sin x + e^x \cos x$, with the given options, we find that it matches option 4.