Chapter 4.5, Problem 6QR

To determine

To find : The line of tangent for the function having equation $f(x)=x{e}^{-x}+1$ .

Answer to Problem 6QR

Explanation of Solution

Given information : The given equation is

  $f(x)=x{e}^{-x}+1$

Where the value of $x=c$ and $c=-1$

Formula used :

Chain Rule: $\frac{d}{dx}\left(f\left(g(x)\right)\right)=\frac{d}{du}\left(f(u)\right)\frac{d}{dx}\left(g(x)\right)$

Product rule: $\frac{d}{dx}\left(f(x)g(x)\right)=\frac{d}{dx}\left(f(x)\right)g(x)+f(x)\frac{d}{dx}\left(g(x)\right)$

Power Rule: $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

Constant Multiply rule: $\frac{d}{dx}\left(cf(x)\right)=c\frac{d}{dx}\left(f(x)\right)$

The given function is $f(x)=x{e}^{-x}+1$

Where $x=c=-1$

The value of the function at the given point: $y_0=f\left(-1\right)=1-e$

The slope of the tangent line at $x=c$ will be the derivative of the function evaluated at $x=c$ :

So, $M\left(c\right)=f\text{'}\left(c\right)$

First find the first derivative of $f(x)=x{e}^{-x}+1$

The derivative obtained for a sum or difference is the sum or difference of derivatives.

So, $\left(\frac{d}{dx}\left(x{e}^{-x}+1\right)\right)=\left(\frac{d}{dx}\left(x{e}^{-x}\right)+\frac{d}{dx}(1)\right)$

It is noted that the derivative of a constant is always zero.

So, $\left(\frac{d}{dx}(1)\right)+\frac{d}{dx}\left(x{e}^{-x}\right)=(0)+\frac{d}{dx}\left(x{e}^{-x}\right)$

On using the product rule with $f\left(x\right)=x$ and $g(x)=e^{-x}$ , it is obtained that

  $\left(\frac{d}{dx}\left(x{e}^{-x}\right)\right)=\left(\frac{d}{dx}(x)e^{-x}+x\frac{d}{dx}\left(e^{-x}\right)\right)$

The function $e^{-x}$ is a combined $f\left(g(x)\right)$ of functions $f(u)=e^u$ and $g(x)=-x$

On applying chain rule

  $x\left(\frac{d}{dx}\left(e^{-x}\right)\right)+e^{-x}\frac{d}{dx}(x)=x\left(\frac{d}{du}\left(e^u\right)\frac{d}{dx}(-x)\right)+e^{-x}\frac{d}{dx}(x)$

The exponential $\frac{d}{du}\left(e^u\right)$ have derivative $e^u$ .

So. $x\left(\frac{d}{du}\left(e^u\right)\right)\frac{d}{dx}(-x)+e^{-x}\frac{d}{dx}(x)=x\left(e^u\right)\frac{d}{dx}(-x)+e^{-x}\frac{d}{dx}(x)$

Returning to the old variable,

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

Now apply power rule with $n=1$ i.e. $\frac{d}{dx}(x)=1$

  $x{e}^{-x}\frac{d}{dx}(-x)+e^{-x}\left(\frac{d}{dx}(x)\right)=x{e}^{-x}\frac{d}{dx}(-x)+e^{-x}(1)$

Now apply constant multiple rule to the above equation with $c=-1$ and $f\left(x\right)=x$

  $-x{e}^{-x}\left(\frac{d}{dx}(x)\right)+e^{-x}=-x{e}^{-x}(1)+e^{-x}$

On simplification

  $-x{e}^{-x}+e^{-x}=(1-x)e^{-x}$

Therefore gives, $\frac{d}{dx}\left(x{e}^{-x}+1\right)=(1-x)e^{-x}$

Hence, $M\left(c\right)=f\text{'}\left(c\right)=(1-c)e^{-c}$

Now, to find the slope of the given point

  $m=M\left(-1\right)=2e$

The equation for the tangent line is denoted as $y-y_{0\text{}}=m(x-x_{0\text{}})$ .

Putting the values, the obtained result is

  $y-\left(1-e\right)=2e\left(x-(-1)\right)$

Or more simply $y=2ex+1+e$

The equation of the tangent line is $y=2ex+1+e$ .