91Ó°ÊÓ

Use the product rule to find the first derivative of \(y=e^x \sin x\). The product rule states that if we have a function \(y = uv\), then the derivative of \(y\) with respect to \(x\) is given by \(y^{\prime} = u^{\prime}v + uv^{\prime}\), where \(u^{\prime}\) and \(v^{\prime}\) are the derivatives of \(u\) and \(v\) with respect to \(x\). In our case, the function \(y=e^x \sin x\) can be written as a product of two functions \(u=e^x\) and \(v=\sin x\). So we first need to find the derivatives of \(u\) and \(v\) with respect to \(x\): $$ u^{\prime}=\frac{d}{dx}e^x=e^x $$ $$ v^{\prime}=\frac{d}{dx}\sin x=\cos x $$ Apply the product rule to find \(y^{\prime}\): $$ y^{\prime}=u^{\prime}v+uv^{\prime}=e^x\sin x + e^x\cos x $$

Now we need to find the second derivative of \(y\). Apply the product rule to the derivative found in step 1 to find \(y^{\prime\prime}\). In this case, the function \(y^{\prime}=e^x\sin x + e^x\cos x\) can be split into the sum of two product functions: \(w=e^x\sin x\) and \(z=e^x\cos x\). We need to find the derivatives of \(w\) and \(z\): $$ w^{\prime}=\frac{d}{dx}(e^x\sin x)=e^x\cos x + e^x\sin x $$ $$ z^{\prime}=\frac{d}{dx}(e^x\cos x)=e^x\cos x - e^x\sin x $$ Now, add \(w^{\prime}\) and \(z^{\prime}\) together to find \(y^{\prime\prime}\): $$ y^{\prime\prime}=w^{\prime}+z^{\prime}=(e^x\cos x + e^x\sin x) + (e^x\cos x - e^x\sin x) $$ Simplify the expression to obtain the final result: $$ y^{\prime\prime}=2e^x\cos x $$ Thus, the second derivative of the given function is \(y^{\prime\prime}=2e^x\cos x\).