Question

Let y=x*e^x. find (d^2 y/dx^2)

Answer 1

We have to find the 2nd derivative of the function:

$y=xe^x$

We would need to know the product rule of derivatives, which is:

If u and v are two functions, then the derivative of u*v is:

$\begin{gathered} \text{If y=uv} \\ dy=uv^{\prime}+vu^{\prime} \end{gathered}$

So, we have:

$\frac{dy}{dx}=x\frac{d}{dx}(e^x)+e^x\frac{d}{dx}(x)$

Remember, the derivative of e^x is e^x and the derivative of "x" is 1.

Thus, we have:

$\begin{gathered} \frac{dy}{dx}=x\frac{d}{dx}(e^x)+e^x\frac{d}{dx}(x) \\ \frac{dy}{dx}=xe^x+e^x(1) \\ \frac{dy}{dx}=xe^x+e^x \end{gathered}$

To get the 2nd derivative, we again have to use the product rule of differentiation on xe^x and just do the differentiation of e^x and sum it. Thus, the process is shown below:

$\begin{gathered} \frac{dy}{dx}=xe^x+e^x \\ \frac{d^2y}{dx^2}=x\frac{d}{dx}(e^x)+e^x\frac{d}{dx}(x)+\frac{d}{dx}(e^x) \\ \frac{d^2y}{dx^2}=xe^x+e^x+e^x \\ \frac{d^2y}{dx^2}=xe^x+2e^x \end{gathered}$

The 2nd derivative of the function shown is:

$\frac{d^2y}{dx^2}=xe^x+2e^x$