Question

Find \(y^{\prime \prime}\) for the following functions. $$y=e^{x} \sin x$$

Short answer

Question: Determine the second derivative (\(y^{\prime \prime}\)) of the function $$y=e^{x} \sin x$$. Answer: The second derivative of the given function is $$y^{\prime \prime} = e^x(2\cos x)$$

Step-by-step solution

Identify the two functions

Let's identify the two functions being multiplied in the given function: $$u=e^x$$ $$v=\sin x$$

Find the first derivative of each function

Next, we will find the first derivative of both u and v: $$u'=e^x$$ $$v'=\cos x$$

Apply the product rule to find the first derivative of y

Now, using the product rule \((uv)' = u'v + uv'\), we can find the first derivative of y: $$y' = (e^x)(\cos x) + (e^x)(\sin x)$$ $$y' = e^x(\cos x + \sin x)$$

Find the second derivative of y

To find the second derivative of y, we will once again use the product rule on the first derivative, treating \(e^x\) as u and \((\cos x + \sin x)\) as v. First, we need to find the derivative of both u and v: $$u'=e^x$$ $$v'=-\sin x + \cos x$$ Now, apply the product rule: $$y^{\prime \prime} = (e^x)(-\sin x + \cos x) + (e^x)(\cos x + \sin x)$$ $$y^{\prime \prime} = e^x(-\sin x + \cos x + \cos x + \sin x)$$

Simplify the expression

Finally, simplify the expression for the second derivative: $$y^{\prime \prime} = e^x(2\cos x)$$