Chapter 3.2, Problem 31E

To determine

To find:

The equation of the tangent line to the curve at the given point

Answer to Problem 31E

The equation of the tangent line is $y=2x^2{e}^x+2x{e}^x$

Explanation of Solution

Given:

The function is

  $y=2x{e}^x,\left(0,0\right)$

Concept used:

An equation of a line $l$ through a point $P_1\left(x_1,y_1\right)$ with slope $m$ . If $P\left(x,y\right)$ is any point with $x\ne x_1$

Then P is on $l$ if and only if the slope of the line through $P_1$ and P

The equation is $y-y_1=m\left(x-x_1\right)$

Calculation:

The function

  $y=2x{e}^x\mathrm{...................}\left(1\right)$

The derivative of a function

  $\begin{array}{l}y=f\left(x\right)\\ y^{\prime }=f^{\prime }\left(x\right)=\frac{dy}{dx}\end{array}$

Differentiating the equation (1) with respect to $x$

  $\begin{array}{l}\frac{dy}{dx}=\frac{d}{dx}\left(2x{e}^x\right)\\ \frac{dy}{dx}=\left(2x\right)\frac{d}{dx}\left(e^x\right)+\left(e^x\right)\frac{d}{dx}\left(2x\right)\\ \frac{dy}{dx}=2x{e}^x+2e^x\\ m=\frac{dy}{dx}=2e^x\left(x+1\right)\end{array}$

The point is $\left(0,0\right)$

The equation is

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y-0=2e^x\left(x+1\right)\left(x-0\right)\\ y=2x^2{e}^x+2x{e}^x\end{array}$